LoRA Probe Activations - Gate, Up, and Down Projections

All projections show: input activations ⋅ A matrix (rank-1 LoRA neuron activations)

Layer 0

GATE_PROJ

Top 16 Positive Activations
Rollout 43, Act: 0.539
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nThere is a C-NOT gate
Rollout 5, Act: 0.539
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapez
Rollout 63, Act: 0.539
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $P$ denote
Rollout 94, Act: 0.539
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nDetermine which set of states mentioned
Rollout 34, Act: 0.539
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind continuous functions $x(t
Rollout 60, Act: 0.539
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA stationary source emits sound of frequency
Rollout 55, Act: 0.539
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet\n\[\mathbf{
Rollout 88, Act: 0.539
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nConsider Convolutional Neural Network D
Rollout 84, Act: 0.539
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGary purchased a large beverage, but
Rollout 47, Act: 0.539
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA train with cross-sectional area $
Rollout 78, Act: 0.539
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nJon and Steve ride their bicycles along
Rollout 11, Act: 0.539
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nSuppose $a,\,
Rollout 12, Act: 0.539
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nProve that the function \[
Rollout 15, Act: 0.539
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $n\geq
Rollout 13, Act: 0.539
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nJoanie takes a $\$6
Rollout 14, Act: 0.539
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA function $f:[0
Top 16 Negative Activations
Rollout 86, Act: -1.989
1.04.\n\nSo case4: ��1.04. While case3 is≈
Rollout 72, Act: -1.990
(sqrt(5) +1)/4 �� (2.236 +1)/4
Rollout 40, Act: -1.993
1421356 +5 ��19.1421356
Rollout 93, Act: -1.994
2=π2 /1007 �� 0.00987, so
Rollout 4, Act: -1.996
.003606)^2 �� (0.0036)^2 =
Rollout 53, Act: -1.999
)= -243/73 �� -3.333\n\nThen x5
Rollout 72, Act: -2.000
-(sqrt(5) +1)/4 �� -0.8090.\n\nTherefore,
Rollout 22, Act: -2.000
see: sqrt(8/(3π)) �� sqrt(8/9.4248
Rollout 53, Act: -2.001
compute:\n\n-243/73 ��-3.328, and -1
Rollout 53, Act: -2.002
= (-1557)/73 ��-21.328\n\nTherefore area
Rollout 86, Act: -2.008
1/6)*(pi/2)^3 �� (1/6)*(3.875
Rollout 69, Act: -2.016
1 + term2 + term3 +... ��0.0348 +0.0
Rollout 17, Act: -2.017
��8π*1.41 ��11.28*3.14
Rollout 86, Act: -2.017
(pi/2)^3 /3 ) ��(1/2)*(3.875
Rollout 17, Act: -2.022
.41 ×10^-33 ��8π*1.41 ��1
Rollout 69, Act: -2.056
3)/1 = 2 +√3 ��3.732, which is not close

UP_PROJ

Top 16 Positive Activations
Rollout 47, Act: 1.331
mass flow rate must be conserved, so �� * A1 * v1 = �� *
Rollout 46, Act: 1.301
1 and M2, each at a distance ��1 and ��2 from the point (x
Rollout 25, Act: 1.297
we assume the asteroid is a sphere with density ��, then mass m = (4/3)
Rollout 47, Act: 1.293
:\n\np0 + (1/2) �� vt2 = p + (1/2)
Rollout 47, Act: 1.281
2 = p + (1/2) �� v_air2\n\nThus, the pressure difference p
Rollout 87, Act: 1.281
a corner of a cube with uniform charge density �� is:\n\nV = (ρ a2)/(4
Rollout 47, Act: 1.280
0 - p = (1/2) �� [( (4/3 vt )2 - vt
Rollout 47, Act: 1.278
- p = (7/(2N)) �� vt2, and we need to find N.
Rollout 46, Act: 1.273
that U is given as such, so perhaps ��1 and ��2 are radial distances from two
Rollout 47, Act: 1.271
0 - p = (1/2) �� (v_air2 - vt2)\n\nWe already
Rollout 47, Act: 1.262
2 ) ] = (1/2) �� (7/9 vt2 ) = (7
Rollout 46, Act: 1.261
problem hasn't specified, maybe we can define ��1 and ��2 as functions of x and
Rollout 46, Act: 1.258
�ρ2/∂x. But If ��1 is the distance from M1 to the point
Rollout 47, Act: 1.257
2 ) = (7/18) �� vt2\n\nNow, comparing this to the given
Rollout 47, Act: 1.256
2 ) ] = (1/2) �� [ (16/9 vt2 -
Rollout 46, Act: 1.255
M1 and M2 are stationary. Then ��1 and ��2 are the distances from the
Top 16 Negative Activations
Rollout 52, Act: -0.456
<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this problem about yummy integers. The question is:
Rollout 26, Act: -0.465
<|im_start|>assistant\n<|im_start|>think\nOkay, let's try to figure this problem out. So, we have
Rollout 18, Act: -0.468
<|im_start|>assistant\n<|im_start|>think\nAlright, let's tackle this problem step by step. First, let me
Rollout 79, Act: -0.470
<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this crossword clue. The clue is "Noticed
Rollout 91, Act: -0.474
<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this geometry problem step by step. The question is
Rollout 45, Act: -0.480
)=8 +16 -4=20 ✔️\n\nf(-4)= -64 +
Rollout 45, Act: -0.488
64 +64 -4= -4 ✔️. Okay, matches.\n\nSo f(x)
Rollout 45, Act: -0.493
(1)=1 +4 -4=1 ✔️\n\nf(2)=8 +16
Rollout 45, Act: -0.496
8+16-4=20, ✔️, f(-4)=-64 +
Rollout 45, Act: -0.499
16=55+4=59 ✔️.\n\nAnother test: x=-1, y
Rollout 45, Act: -0.499
(1)=1+4-4=1 ✔️, f(2)=8+16
Rollout 45, Act: -0.507
-64 +64 -4=-4 ✔️. Also f(0)=0 +0
Rollout 45, Act: -0.519
(0)=0 +0 -4=-4 ✔️. So works.\n\nBut to make sure that
Rollout 45, Act: -0.522
+12)= -2 +6=4 ✔️.\n\nYes, matches. So function seems okay
Rollout 23, Act: -0.528
<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this problem step by step. So, we have
Rollout 54, Act: -0.541
<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this question step by step. First, I need

DOWN_PROJ

Top 16 Positive Activations
Rollout 49, Act: 1.277
Prove that if every subspace of a Hausdorff space $X$ is
Rollout 49, Act: 1.258
's unpack the terms here.\n\nFirst, a Hausdorff space is a topological space where any
Rollout 49, Act: 1.257
statement is: If every subspace of a Hausdorff space X is σ-compact, then
Rollout 49, Act: 1.251
. Therefore, if we can show that a Hausdorff space where every subset is Fσ must
Rollout 49, Act: 1.236
is closed. But a perfect set in a Hausdorff space is uncountable. Maybe that
Rollout 49, Act: 1.232
of closed sets. But wait, in a Hausdorff space, a σ-compact subspace
Rollout 57, Act: 1.224
concepts. Its properties include being completely regular, Hausdorff, but not normal. \n\nSince the
Rollout 49, Act: 1.221
instance, in the real line (which is Hausdorff, uncountable), there are subsets
Rollout 49, Act: 1.220
here.\n\nWait, but our space X is Hausdorff, not necessarily metrizable. Hmm
Rollout 49, Act: 1.220
be: If X is uncountable and Hausdorff, then there is a subspace that
Rollout 49, Act: 1.219
subset is σ-compact, then in a Hausdorff space, that implies that every subset is
Rollout 49, Act: 1.219
σ-compact by hypothesis. But in a Hausdorff space, perfect sets are closed, hence
Rollout 49, Act: 1.218
In such a space, consider X, a Hausdorff space where every subset is Fσ.
Rollout 49, Act: 1.218
, a common example of an uncountable Hausdorff space that isn't σ-compact is
Rollout 49, Act: 1.218
X is countable.\n\nWait, in a Hausdorff space where singletons are Gδ,
Rollout 49, Act: 1.215
for a theorem. I recall that if a Hausdorff space is hereditarily Lindelö
Top 16 Negative Activations
Rollout 44, Act: -0.914
that case, irreversible work is higher.\n\nBut in another example where initial p2=1.5 atm
Rollout 40, Act: -0.915
=144, remainder 6. So ��^150=ω^(8*
Rollout 1, Act: -0.917
x' - y⟩ = 0.\n\nSo another way to think of this is that the set {
Rollout 21, Act: -0.919
thought before, with only finite expressions allowed. So another approach: can we create an equation that must hold
Rollout 62, Act: -0.920
another number has three prime factors. So even if another number is semiprime (two primes)
Rollout 57, Act: -0.921
to analyze the topology generated by Lebesgue-measurable sets containing a point \( x \) with density
Rollout 41, Act: -0.921
my previous reasoning. Let's see.\n\nBut perhaps another way is using combinations and permutations.\n\nHmm. Wait
Rollout 41, Act: -0.922
. So we need to handle this carefully.\n\nSo another way is:\n\nFor three-digit numbers with exactly two
Rollout 57, Act: -0.923
neighborhood basis consists of all such Lebesgue-measurable sets with density 1 at x. So,
Rollout 93, Act: -0.930
to different powers. Hmm.\n\nBut maybe there's another approach. For integrals of cosine raised to high
Rollout 20, Act: -0.938
1) ��1 mod m.\n\nThen following similar steps as before: multiplying by (p-1
Rollout 51, Act: -0.945
is even around pi, let's see.\n\nBut another view: product of sin8x and |sin
Rollout 57, Act: -0.965
the family of all density neighborhoods at x (measurable sets with density 1 at x) form a
Rollout 8, Act: -0.969
10P5. Yeah.\n\nBut perhaps another perspective: if she's selecting 5 cards from
Rollout 57, Act: -0.974
measure one. Such sets are called saturated nonmeasurable sets, but are typically only possible under certain set
Rollout 81, Act: -1.014
at A, edges AB and AC. Sotriangle area can also be written as:\n\n Area=1

Layer 1

GATE_PROJ

Top 16 Positive Activations
Rollout 49, Act: 2.323
orff) and it is a Hereditarilyseparable HS space, but I don't think we
Rollout 33, Act: 2.270
.\n\nk=3: permutations with first three permutationof1-3: 6*1=6
Rollout 54, Act: 2.248
iles (1. 4-methylcyclohexan-1-olate, 2.
Rollout 25, Act: 2.247
ola with perigee r_p and the apse line is where the closest approach occurs. The parameters
Rollout 42, Act: 2.236
final product produced when cyclobutyl(cyclopropyl)methanol reacts with phosphoric acid
Rollout 57, Act: 2.229
Wait, perhaps a method used in proving Urysohn's lemma. But in a normal space,
Rollout 44, Act: 2.227
reversible process, since the external pressure is always infinitesimally larger than the internal pressure. If the
Rollout 42, Act: 2.178
the product formed when cyclobutyl(cyclopropyl)methanol reacts with phosphoric acid
Rollout 24, Act: 2.167
.\n<|im_start|>user\nConsider any rectangular table having finitely many rows and columns, with a real \n\n
Rollout 10, Act: 2.154
you add to S, you must exclude only finitely many others (i.e., n +s
Rollout 46, Act: 2.154
as (acceleration_gravity + acceleration_centrifugal). Because:\n\nFor each term:\n\n∇(-
Rollout 57, Act: 2.149
lemma. But in a normal space, Urysohn's lemma holds. However, this space isn
Rollout 65, Act: 2.146
solution.\n\n But hold on, solution from math connoisseurs shows that acceptable similar prisms require harmonic
Rollout 87, Act: 2.143
after all this, if I cannot proceed further analytically, and the problem likely expects using the integral
Rollout 37, Act: 2.142
015! For any integer substitution, needto see its Coefficients are divisible by 2
Rollout 49, Act: 2.141
uncountable and hereditariy Lindelof, then it has a uncountablesubset with
Top 16 Negative Activations
Rollout 86, Act: -1.587
is 1 -0=1.\n\nThus, integral of x sin x dx is 1. Therefore
Rollout 51, Act: -1.587
3 = -4/63\n\nTherefore, Integral B = -4/63\n\nThus,
Rollout 86, Act: -1.587
1/2)cos2x.\n\nThus, integral= - [ - (1/2)x cos
Rollout 86, Act: -1.591
2 dx, v=sin x.\n\nThus, integral = x3 sin x | from 0 to
Rollout 51, Act: -1.598
, leading to -4/63.\n\nIn Integral B steps:\n\n sin9x - sin7x
Rollout 86, Act: -1.600
1/2) sin2x.\n\nThus, integral becomes (x2)(1/2) sin
Rollout 86, Act: -1.602
1/2) sin2x.\n\nThus, integral= x2*(1/2 sin2x
Rollout 86, Act: -1.616
5.\n\nThus, plug back into case6:\n\n integral of x^2 cos^2x dx=(
Rollout 51, Act: -1.650
π} cos8t sint dt\n\nTherefore, Integral B becomes:\n\n∫0^π cos8x
Rollout 51, Act: -1.661
sin7x] / 2\n\nTherefore, Integral B over 0 to π is:\n\n2 *
Rollout 51, Act: -1.664
cos9x] / 2\n\nTherefore, Integral A over 0 to π is ��0
Rollout 51, Act: -1.670
| dx = -4/63\n\nSo Integral B = -4/63.\n\nSo yes
Rollout 51, Act: -1.672
sinx| = -sinx.\n\nTherefore, Integral A can be written as:\n\n∫0^π
Rollout 87, Act: -1.676
/ sqrt(x2 + y2)).\n\nThen Integral becomes:\n\nV = (ρ/(4 π ε
Rollout 51, Act: -1.687
8x |sinx| dx\n\nSimilarly, Integral B: ��02π cos8x
Rollout 51, Act: -1.701
-cos(nx)/n + C\n\nSo, Integral B becomes:\n\n[-cos9x/9 +

UP_PROJ

Top 16 Positive Activations
Rollout 25, Act: 1.391
*(6.67e-11 * m )/v2\n\nSolving for m,\n\n
Rollout 25, Act: 1.368
=6.849e14→ M_e +m=6.849e
Rollout 22, Act: 1.349
(8 * R * T / (π * M)) = sqrt(8 * 8.3
Rollout 27, Act: 1.348
6 kg * (1.49e5 m/s)^2) / (3 * 1
Rollout 22, Act: 1.327
(8 * R * T / (π * M)).\n\nR = 8.314
Rollout 27, Act: 1.327
times 10^{5} \text{ m/s}}$ and the estimated critical temperature $T
Rollout 22, Act: 1.326
(8 * R * T / (π * M)), where\nR=8.314
Rollout 27, Act: 1.326
times 10^{5} \text{ m/s}}\) and the estimated critical temperature \( T
Rollout 25, Act: 1.323
�� 1.094e6 m\n\nDenominator:\n\nsqrt(1 + ( (
Rollout 3, Act: 1.307
as \boxed{169 \text{ m/s}}.\n\nHowever, depending on source of problem
Rollout 27, Act: 1.307
� 1.4915e5 m/s. So approx 1.49e
Rollout 27, Act: 1.304
�� 1.491e5 m/s.\n\nYes, so approximately 1.4
Rollout 27, Act: 1.301
2.2201e10 m2/s2\n\nMultiply by mass:\n\n1.
Rollout 66, Act: 1.296
, m2 -n2 ) = gcd( m2 +n2, (m2 +n
Rollout 66, Act: 1.289
m2 -n2 ) ) = gcd( m2 +n2, 2n2 )
Rollout 25, Act: 1.288
2 = 86.49e6 m2/s2\n\nNumerator: 10
Top 16 Negative Activations
Rollout 70, Act: -2.066
0. Then using this, let's proceed.\n\nThus:\n\nEquation:314d3 +
Rollout 83, Act: -2.070
is a symmetry that cycles the three residue groups.\n\nThus,\n\nThe rotational group is of the nonagon acting
Rollout 65, Act: -2.071
leads to cube. So only this solution exists.\n\nThus, this approach also yields cube. So cutting along
Rollout 27, Act: -2.073
density. However, problem mentions no such number.\n\nThus possibly, in absence of more details, proceeding with
Rollout 20, Act: -2.073
for 2, which appears once in both.\n\nThus LCM is 2*3*7*
Rollout 68, Act: -2.076
is in the domain. Hence, one solution.\n\nThus, in general, for any k<0,
Rollout 35, Act: -2.078
418, which is what was required.\n\nThus, this seems like the logical conclusion, and the
Rollout 1, Act: -2.084
with countably or uncountably many vertices.\n\nIndeed, in an infinite-dimensional Hilbert space, one
Rollout 96, Act: -2.086
the original positive/right/east in original coordinates.\n\nThus, point P(-14,27)
Rollout 67, Act: -2.089
So sequence peaks around there, and then decreases.\n\nThus, the term n=3 is 18
Rollout 83, Act: -2.091
by rotation number3, which cycles the groups.\n\nThus, permutation of (group0, group1,
Rollout 57, Act: -2.092
However, this is similar to our previous arguments.\n\nIndeed, if take P = irrationals, then
Rollout 67, Act: -2.101
the eighth term remains consistent with the quadratic model.\n\nThus, the eighth term of the sequence is \(\
Rollout 12, Act: -2.102
or incomplete elliptic integral. Let's check.\n\nIndeed, in the integral:\n\n∫_{ν}^
Rollout 21, Act: -2.140
statement requires equations only (E=0). Inequalities aren't allowed as they don't fit into
Rollout 73, Act: -2.162
. So this verification also gives the same result.\n\nThus, the sum N is 1639

DOWN_PROJ

Top 16 Positive Activations
Rollout 79, Act: 0.666
't make sense). No.\n\nAlternatively, "Organism Hen". No, not making sense.\n\nAlternatively,
Rollout 98, Act: 0.613
million years. The white dwarf would be the remnant after the star has shed its envelope. Now,
Rollout 98, Act: 0.600
Wait, wait. A white dwarf is the remnant of a star that has shed its outer layers.
Rollout 98, Act: 0.600
A0V. The white dwarf is the remnant of a B5V star. Since massive stars
Rollout 98, Act: 0.595
a WD. So if the WD is the remnant of a B5 main sequence star (mass ~
Rollout 98, Act: 0.512
Wait, actually, the initial mass leading to WD remnants is usually roughly up to 8Msun.
Rollout 98, Act: 0.511
. If DA4 is a white dwarf (remnant of a star) and L4 is a brown
Rollout 65, Act: 0.486
.\n\n But hold on, solution from math connoisseurs shows that acceptable similar prisms require harmonic mean
Rollout 57, Act: 0.471
P和P’是开的。若他们是互不相交,能否用开集将Q
Rollout 29, Act: 0.452
the bottom of the jug (main part) should normally be rho*g*h. However, if the handle
Rollout 29, Act: 0.446
h is the depth below the surface. So, normally, if you have a container open to the atmosphere
Rollout 4, Act: 0.446
2.6% total spread. But since normally uncertainty is defined as ± half the total spread,
Rollout 80, Act: 0.441
First, the problem states that neutrinos, which normally rarely interact, are now being completely absorbed by the
Rollout 29, Act: 0.439
a manufacturing defect". So the handle, which is normally open (so that when you pour, milk can
Rollout 99, Act: 0.427
573.\n\nBut since after two iterations, normally would to compute x4, as I earlier showed
Rollout 15, Act: 0.426
if exactly one neighbor is 1 (counted normally), but consider two neighbors: "exactly one
Top 16 Negative Activations
Rollout 97, Act: -0.547
since we're dealing with pressure, maybe using pressure instead of concentration. Since in a gas phase reaction at
Rollout 30, Act: -0.550
relations, that might be challenging.\n\nWait, perhaps instead of trying to find explicit expressions for a_n and
Rollout 22, Act: -0.551
there's another formula if we use the gas constant instead of Boltzmann's constant. Let's check
Rollout 67, Act: -0.551
2 + Bn + C. Thus, maybe instead of solving for the variables a, d, p
Rollout 42, Act: -0.556
ocation. Let's consider that.\n\nAlternatively, perhaps instead of a carbocation rearrangement, the elimination occurs
Rollout 81, Act: -0.558
/7≈3.14, so negative instead.\n\nThus,the ambiguity arises because the angle computed via
Rollout 99, Act: -0.558
question says "decimal value", does it want decimal instead of fraction? But given the question asks for decimal
Rollout 28, Act: -0.560
why all these conics must be hyperbolas instead of other types like ellipses or parab
Rollout 23, Act: -0.563
carefully.\n\nBut since this step is critical, maybe instead, approximate ln(A). For answer D,
Rollout 23, Act: -0.564
.\n\nAh! If someone mistakenly computed the Total activity instead of just alpha activity, they would get Total activity
Rollout 16, Act: -0.570
in this case, it's a bit different because instead of collecting one coupon at a time, each sap
Rollout 11, Act: -0.585
Altizio, which suggests there is a clever solution instead of heavy computation.\n\nRecall that from abc =
Rollout 86, Act: -0.588
sinx and cosx multiplicative terms as possible instead of x.\n\nBecause sinx and cosx are
Rollout 99, Act: -0.591
ant method but uses a quadratic interpolation through three points instead of a linear one. This allows it to potentially
Rollout 74, Act: -0.600
." Maybe the model here is considering permanent magnetism instead of induced. Permanent ship's magnetism would produce
Rollout 50, Act: -0.609
some way. Maybe we should find a better approach instead of expanding all this.\n\nAlternatively, perhaps start over

Layer 2

GATE_PROJ

Top 16 Positive Activations
Rollout 37, Act: 1.251
[\n[a_1^{k_1} a_1^{k_2} ...
Rollout 99, Act: 1.208
c)/(b ± sqrt(b2 - 4 a c))\n\nSo that's the formula. Let's
Rollout 99, Act: 1.161
c)/(b ± sqrt(b2 - 4 a c))\n\nFirst, compute discriminant D = b
Rollout 37, Act: 1.126
]\n[a_2^{k_1} a_2^{k_2} ...
Rollout 37, Act: 1.088
015}^{k_1} a_{2015}^{k_2
Rollout 11, Act: 1.029
= |b|^2*(|b|^4 a ) ) = |b|^6 a → Therefore
Rollout 40, Act: 1.020
_{k=0}^{n-1} a^k ). Then using geometric series formula: (
Rollout 30, Act: 1.016
1}] / [a_{n+1} a_{n+3}]. Then, the term
Rollout 17, Act: 1.016
{r})=\frac{e^{-r / a_{0}}}{\sqrt{\pi a_{
Rollout 40, Act: 1.015
_{k=0}^{n-1} a^k}. This is assuming a ≠ 1
Rollout 12, Act: 1.011
1 / sqrt( (x^2 - a^2)(b^2 - x^2
Rollout 14, Act: 1.001
) ]3 +3 [sqrt(2 K a )]^2 (a/2 ) +3
Rollout 30, Act: 0.998
∞ 1/( (a_m -1 ) a_m ). But that's the same as the original
Rollout 14, Act: 0.995
/2 ) +3 (sqrt(2 K a ))(a2/4 ) +a3
Rollout 30, Act: 0.986
= 1/( (a_m -1 ) a_m )\n\nWhich suggests that the sum can be rewritten
Rollout 30, Act: 0.979
first sum into 1 + sum [1/( a_k(a_k -1 ) ]\n\nThe second sum
Top 16 Negative Activations
Rollout 56, Act: -2.765
+1})√3. So as n approaches infinity, S=√3. Therefore, the suprem
Rollout 81, Act: -2.767
absolute value gives22/7. In the denominator case h2 -51<0, so
Rollout 99, Act: -2.771
� 2.657\n\nSo the denominator is b ± sqrt(D). Which is -0
Rollout 51, Act: -2.793
/9 -2/7\n\nConvert to common denominator:\n\n2*(7 -9)/63 =
Rollout 90, Act: -2.802
7-digit number, which must start with a non-zero digit, hence N must be from 10
Rollout 53, Act: -2.806
29/292:\n\nConvert to common denominator:\n\n81/4= (81 *
Rollout 93, Act: -2.818
)}{1007}}\) is non-zero only when \(1007 - j\)
Rollout 16, Act: -2.827
192/31: convert to common denominator 7*31=217:
Rollout 16, Act: -2.829
before concluding, double checking the calculation step where common denominator was 1953, so:\n\n1
Rollout 16, Act: -2.856
So writing all in terms of fractions with a common denominator (though maybe we can add them step by step
Rollout 81, Act: -2.861
>0, so angle is acute.\n\nBut in denominator [1 - tan alpha tan beta], as per
Rollout 16, Act: -2.864
64/63\n\nConvert to common denominator. The denominators are 217 and
Rollout 16, Act: -2.880
-64/63: convert to common denominator 217*63=13
Rollout 99, Act: -2.895
6431374\n\nThen the denominator is:\n\nOption1: b + sqrt(D)=
Rollout 16, Act: -2.919
-1/63.\n\nThen converting to common denominator 1953:\n\n11/2
Rollout 16, Act: -2.932
}\n\]\n\nConverting all terms to a common denominator and simplifying, we find the expected number of

UP_PROJ

Top 16 Positive Activations
Rollout 20, Act: 1.863
m, we need either:\n\n- If p ��1 mod m: then kn ��0 mod
Rollout 93, Act: 1.848
1007.\n\nThen:\n\nIf m ��0 mod1007, then sum_{
Rollout 2, Act: 1.837
We need to solve for x in:\n\nx �� 109 mod125\n\nx
Rollout 20, Act: 1.829
��0 mod m\n\n- If p �� other residues mod m: p^{kn} �
Rollout 20, Act: 1.828
43.\n\nSubcases:\n\n- If p ��1 mod43: sum is kn+1
Rollout 20, Act: 1.824
43.\n\nFirst, note that if p �� 0 mod 43, then p =
Rollout 2, Act: 1.818
��5 mod8.\n\nTherefore, P �� 5 mod8.\n\nAlright, now onto modulo
Rollout 10, Act: 1.783
3 mod4:\n\nPossible moves:\n\nIf n ��0 mod4: subtract0 or1.\n\n subtract
Rollout 10, Act: 1.783
;0 -1=3;\n\nIf n ��1 mod4: subtract1 or0.\n\nn
Rollout 20, Act: 1.779
=kn +1. Therefore, if p ��1 mod m, we need kn +1 �
Rollout 93, Act: 1.773
so sum is N. Thus:\n\nIF m ��0 mod N => sum_{k=1}
Rollout 20, Act: 1.769
+1 mod m. So, if p ��1 mod m, then σ(p^{kn})
Rollout 20, Act: 1.766
+ ... + p^{kn}. If p ��1 mod m (meaning m is 43
Rollout 90, Act: 1.759
inclusive, we can express this as:\n\nM �� -9N mod 1009\n\n
Rollout 2, Act: 1.758
P mod 8.\n\nSince 10 �� 2 mod 8, so 10
Rollout 70, Act: 1.753
.\n\nWait, is that correct?\n\nBecause N ��d0 mod7 (since N= ... +
Top 16 Negative Activations
Rollout 42, Act: -1.647
three substituents: one hydroxyl (from the alcohol), a cyclobutyl ring,
Rollout 26, Act: -1.651
four sig figs), the Enthalpy of formation of Hg(g) is given as 6
Rollout 98, Act: -1.655
.85Msun), K3V, A star with a mass of 0.9Ms
Rollout 79, Act: -1.658
A common term that means investigators' facility is "Operations Center". Let's check. Operations = 1
Rollout 7, Act: -1.660
56*(N/3)-42*( (N/3) -x).\n\nSimplify:\n\n
Rollout 56, Act: -1.664
esyu starts with a right-angled triangle \( A_0A_1A_2 \)
Rollout 25, Act: -1.664
= 3175e3 m (from previous calcs), then r < r_p_final
Rollout 25, Act: -1.667
= 3175e3 m (from earlier calculation)\n\nr_p_final = 63
Rollout 25, Act: -1.684
080,625e6 (from earlier, when we calculated r_p_initial)\n(
Rollout 98, Act: -1.699
when the B5 star becomes a WD. Binary interactions or mass loss scenarios allow for plausible coexistence.\n\n
Rollout 98, Act: -1.701
and **M4V** (red dwarf, Population I) stars are younger and metal-rich. These
Rollout 25, Act: -1.727
1.008e13 * (9300)^4 ) / (7
Rollout 98, Act: -1.731
longer lifetime. The W Virginis star is a Population II Cepheid, which indicates it's older
Rollout 98, Act: -1.753
while **G2V** (Sun-like, Population I) and **M4V** (red
Rollout 21, Act: -1.793
with variables x1, x2, etc., applications of f, addition, multiplication, subtraction, and
Rollout 98, Act: -1.846
is in a system with a G2V (Population I), would that be possible? Maybe in the

DOWN_PROJ

Top 16 Positive Activations
Rollout 89, Act: 1.522
problem is altered, thus:\n\nSuppose there's misprint in the problem. Instead 2/5
Rollout 99, Act: 1.515
even with the two iterations, but due to possible misconfusion in the problem statement.\n\nAlternatively, maybe
Rollout 84, Act: 1.482
/9 as much beverage".\n\nAlternatively, someone could misread it as purchasing half as much, and then
Rollout 84, Act: 1.469
as much and drunk twice as much". Some might misinterpret "twice as much" as twice the
Rollout 95, Act: 1.419
not equal. Therefore, the problem is flawed or misinterpretation is there.\n\n Wait, but problem states
Rollout 84, Act: 1.414
time through the problem to make sure I didn't misinterpret anything.\n\nThe problem says "he had purchased
Rollout 95, Act: 1.401
Area ABC, which are not constants.\n\n Unless some miscomputation i made.\n\n Let's recheck area
Rollout 89, Act: 1.374
/5th regardless of boys.\n\nWait, that misinterpretation.\n\nWait, If someone reads it as
Rollout 64, Act: 1.372
be like that.\n\nPerhaps there's a mistake in misapplying the initial formula.\n\nHold on. Let
Rollout 89, Act: 1.360
a whole number.\n\nUnless the test question has a misprint, and actually there are 0.2
Rollout 30, Act: 1.355
step of the reasoning because mistakes in index shifts or misapplying the recursions could affect the answer.\n\n
Rollout 89, Act: 1.351
to interpret this?\n\nWait, maybe the problem was misstated.\n\nAlternatively, maybe the original statement is
Rollout 41, Act: 1.340
, right?\n\nWait, hold on. Maybe I misread. The question is "at most two different
Rollout 83, Act: 1.329
would fix the arrangements, since other rotations would cause misalignment.\n\nBut wait, but maybe this is incorrect
Rollout 89, Act: 1.288
th of children are girls." That could be a misinterpretation. If "2/5th of
Rollout 99, Act: 1.276
x4. However, likely, there was a miscount in the iteration steps.\n\nAlternatively, perhaps Mull
Top 16 Negative Activations
Rollout 84, Act: -1.048
drank m/n of 1 unit, so the actual amount he drank originally was m/n * 1
Rollout 37, Act: -1.052
2015!. Then perhaps through combinatorial interpretations.\n\nWait, hold on. Let me consider
Rollout 48, Act: -1.055
bytes worth of header. Then, presumably, the actual IP packet with data has 52 bytes total
Rollout 37, Act: -1.057
consecutively assigned. Or through combinatorial interpretations. But I haven't seen this before.
Rollout 24, Act: -1.062
to a kind of fixed point in some combinatorial structure.\n\nAnother idea: Suppose (R1,
Rollout 93, Act: -1.067
remember that such sums can be evaluated using combinatorial identities or orthogonality relations with Cheb
Rollout 24, Act: -1.073
variants like rank.\n\nAlternatively, a pivotal combinatorial lemma.\n\nAlright, perhaps it's finally time to
Rollout 33, Act: -1.074
maybe there's a generating function or some combinatorial trick to handle it.\n\nAlternatively, perhaps we can
Rollout 16, Act: -1.079
we can use inclusion-exclusion or some combinatorial approach.\n\nExpanding the term [1 - (
Rollout 26, Act: -1.091
absolute terms, Δ(n)*R*T. The actual value computed was -7.4202
Rollout 48, Act: -1.092
would be 0xFB12. But the actual value would then be FB12. But if
Rollout 41, Act: -1.099
two different digits?\n\nThis is a classic combinatorial question.\n\nNumber of such numbers = C(1
Rollout 24, Act: -1.105
there probably is a theorem for this in combinatorial game theory or hypergraphs that I don't remember
Rollout 26, Act: -1.110
= -180.76 (precise), but given possible uncertainties:\n\nIf we consider all
Rollout 81, Act: -1.112
different equations. However, this is not possible—the actual mathematical relationship must consistent. So I need to figure
Rollout 10, Act: -1.154
called losing positions or P-positions in combinatorial game theory. The problem states there are infinitely many

Layer 3

GATE_PROJ

Top 16 Positive Activations
Rollout 88, Act: 0.544
is \(\boxed{100}\).<|im_end|>
Rollout 91, Act: 0.505
|\) is \(\boxed{2}\).<|im_end|>
Rollout 31, Act: 0.499
-\sqrt{5}}{2}}\).<|im_end|>
Rollout 99, Act: 0.492
boxed{2.157} \).<|im_end|>
Rollout 32, Act: 0.479
\(\boxed{4018}\).<|im_end|>
Rollout 0, Act: 0.464
is \(\boxed{128}\).<|im_end|>
Rollout 83, Act: 0.426
is \(\boxed{144}\).<|im_end|>
Rollout 10, Act: 0.391
where the second player can force a win.}<|im_end|>
Rollout 27, Act: 0.343
^{6} \text{ K}}\).<|im_end|>
Rollout 95, Act: 0.328
rational length is \(\boxed{0}\).<|im_end|>
Rollout 53, Act: 0.312
dfrac{27}{2}}\).<|im_end|>
Rollout 1, Act: 0.297
y \text{ exists as described}}\n\]<|im_end|>
Rollout 65, Act: 0.289
) is \(\boxed{40}\).<|im_end|>
Rollout 72, Act: 0.284
{\text{All even positive integers}}\n\]<|im_end|>
Rollout 67, Act: 0.275
is \(\boxed{348}\).<|im_end|>
Rollout 94, Act: 0.265
correct answer is \(\boxed{B}\).<|im_end|>
Top 16 Negative Activations
Rollout 48, Act: -5.014
, then take the one's complement.\n\nBut let's try this step by step:\n\nAdd all the
Rollout 90, Act: -5.019
relate N, L, and M.\n\nBut let's start with 9N + M ��
Rollout 40, Act: -5.021
- ��^{150})\n\nBut let's first compute ��^{150}. Since
Rollout 56, Act: -5.024
1, A2, A3?\n\nBut let's make sure: the original triangle was A0A
Rollout 91, Act: -5.031
= -1. So, position.\n\nBut let's define coordinates. Let's take D as the origin
Rollout 39, Act: -5.033
terms here. Hmm, okay.\n\nSo, let's see. If I need to compute the sum of
Rollout 39, Act: -5.034
7101.\n\nWait, but let's think about it. 10^n - k
Rollout 17, Act: -5.043
0.0254 Tesla.\n\nBut let's also confirm calculations:\n\nGiven:\n\nμ0=4
Rollout 35, Act: -5.043
is a divisor of 10.\n\nBut let's list all possible divisors of 10:
Rollout 1, Act: -5.043
could be uncountable. Hmm.\n\nBut let's think: Suppose that S is a set where all
Rollout 78, Act: -5.046
train to go past Steve.\n\nWait, but let's take this step by step.\n\nFirst, let's
Rollout 99, Act: -5.052
≈2.1573.\n\nBut let's check if perhaps my initial assumption of coefficients is wrong
Rollout 98, Act: -5.059
. The answer is B (3).\n\nBut let's just check for system 1 once more. If
Rollout 35, Act: -5.059
to determine n1 and n2.\n\nBut let's dive a bit deeper. Since Q(n) =
Rollout 23, Act: -5.088
so yes, most have decayed.\n\nBut let's double-check. For e^-73.0
Rollout 52, Act: -5.114
negative. But we need to see.\n\nBut let's take the next step.\n\nFirst, factorizing

UP_PROJ

Top 16 Positive Activations
Rollout 28, Act: 1.236
(a,0), B(0,b), G(a/3, b/3). Let's take
Rollout 9, Act: 1.184
common to triangle $GEM$ and square $AIME$ is $80$ square units.
Rollout 12, Act: 1.177
) \cdot b'(\nu) - F(a(\nu), \nu) \cdot a'
Rollout 28, Act: 1.154
Now, points O(0,0), A(a,0), B(0,b), G(a
Rollout 28, Act: 1.132
(a,0), B(0,b), G(a/3, b/3 ).\n\nThis is the
Rollout 77, Act: 1.131
= [(a + b)/b] * f(a, b - a) ... Wait, hold on
Rollout 59, Act: 1.113
coins must be tails. So the probability is P(A=T)P(B=T)P(C=T)
Rollout 96, Act: 1.112
5}{12}$ that contains the point $A = (24,-1)$ , and let
Rollout 56, Act: 1.106
_{0}$ and the new pasture is triangle $A_{1} A_{2} A_{3
Rollout 28, Act: 1.105
system, with O(0,0), A(a,0), B(0,b), centered at
Rollout 30, Act: 1.105
satisfy $a_{n+1}=\left(a_{n-1}-1\right)\left
Rollout 28, Act: 1.104
In original coordinates O(0,0), A(a,0), B(0,b), G(a
Rollout 77, Act: 1.102
- a) = (b - a) f(a, a + (b - a)) => b
Rollout 56, Act: 1.090
n} A_{n-2}$ to $A_{n+1}$ such that $A_{
Rollout 77, Act: 1.087
= [b / (b - a)] f(a, b - a)\n\nBut this still involves f
Rollout 77, Act: 1.082
- a) = (b - a) f(a, b)\n\nTherefore:\n\nf(a, b)
Top 16 Negative Activations
Rollout 79, Act: -2.075
But not. As before.\n\nWait! A common term that means investigators' facility is "Operations Center".
Rollout 25, Act: -2.075
. Thus, with very large μ_final, perigee can be made as small as desired. However
Rollout 25, Act: -2.076
final}} \)**:\n Using the desired perigee distance \( r_p_{\text{final
Rollout 25, Act: -2.076
\( r \)**:\n Using the original perigee distance \( r_p_{\text{initial
Rollout 56, Act: -2.099
step adds half of the prior amount.\n\nSo general term: area S_n=K_n= (sum
Rollout 65, Act: -2.103
, which is same as original.\n\nAlternatively, non-corresponding ratios. If the new prism has sides
Rollout 46, Act: -2.124
hold on. Let me clarify:\n\nThe cross product term �� × v, in components assuming ��
Rollout 10, Act: -2.224
5, the Grundy numbers may be non-periodic.\n\nBut since manual calculation is tedious, let's
Rollout 25, Act: -2.239
, raising mu further seems possible.\n\nBut as perigee formula:\n\nr_p = [h2/m
Rollout 10, Act: -2.252
we can find a different pattern or a non-periodic structure.\n\nStarting from n=0 (L),
Rollout 83, Act: -2.294
fixed; since the digits are unique, the automorphisms are only rotations, and any rotated copy differs
Rollout 15, Act: -2.306
matrix with 1s on the sub and superdiagonals.\n\nFor instance, if you have this
Rollout 10, Act: -2.311
) were computed for several positions, showing non-periodic but structured behavior.\n\n4. **Constructing Infinite
Rollout 15, Act: -2.325
would have a single 1 on the first superdiagonal and first subdiagonal.\n\nWait, example
Rollout 87, Act: -2.540
5\n\nOther combinations. For instance, considering hypergeometric functions or similar, there is no closed form
Rollout 12, Act: -2.588
. Not exactly matching here.\n\nAlternatively, consider hypergeometric substitutions.\n\nWait, but maybe that's getting

DOWN_PROJ

Top 16 Positive Activations
Rollout 35, Act: 1.497
= \boxed{418}\n\]<|im_end|>
Rollout 31, Act: 1.480
-\sqrt{5}}{2}}\).<|im_end|>
Rollout 19, Act: 1.474
\[\n\boxed{78}\n\]<|im_end|>
Rollout 72, Act: 1.465
{\text{All even positive integers}}\n\]<|im_end|>
Rollout 59, Act: 1.448
[\n\boxed{515}\n\]<|im_end|>
Rollout 5, Act: 1.396
[\n\boxed{181}\n\]<|im_end|>
Rollout 81, Act: 1.368
(angle BAC)=22/7 Therefore, despite formula in second approach appearsConflicting, the coordinate
Rollout 50, Act: 1.365
[\n\boxed{112}\n\]<|im_end|>
Rollout 69, Act: 1.362
[\n\boxed{204}\n\]<|im_end|>
Rollout 60, Act: 1.350
]\n\n\[\n\boxed{6}\n\]<|im_end|>
Rollout 99, Act: 1.350
boxed{2.157} \).<|im_end|>
Rollout 91, Act: 1.349
|\) is \(\boxed{2}\).<|im_end|>
Rollout 80, Act: 1.339
\[\n\boxed{11}\n\]<|im_end|>
Rollout 76, Act: 1.338
[\n\boxed{929}\n\]<|im_end|>
Rollout 68, Act: 1.335
[\n\boxed{501}\n\]<|im_end|>
Rollout 16, Act: 1.329
{789953}\n\]<|im_end|>
Top 16 Negative Activations
Rollout 5, Act: -1.250
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapezoid is
Rollout 34, Act: -1.250
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind continuous functions $x(t),\
Rollout 47, Act: -1.250
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA train with cross-sectional area $S_{
Rollout 43, Act: -1.250
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nThere is a C-NOT gate where the
Rollout 60, Act: -1.250
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA stationary source emits sound of frequency $f
Rollout 55, Act: -1.250
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet\n\[\mathbf{A}
Rollout 63, Act: -1.250
system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $P$ denote the set
Rollout 26, Act: -1.286
s, so given conflicting rules.\n\nHowever, in most such problems, even with intermediate precision, the answer
Rollout 23, Act: -1.289
,可能會選擇D.\n\nBut ��據題目,答案不应该是total,而是alpha的。
Rollout 57, Act: -1.296
of \( \complement U \) (denoted \( V \)). Every point in \( C \
Rollout 11, Act: -1.312
of the form t +1/t.\n\nThus, denote t1 =a, t2= a z
Rollout 57, Act: -1.375
**: The complement of \( U \) (denoted \( \complement U \)) is closed,
Rollout 25, Act: -1.467
be smaller, meaning it could approach closer before breaking apart. However, the question is about making contact with
Rollout 11, Act: -1.873
not on unit circle either.\n\nAdditionally, (denote S = a + b + c, T =
Rollout 24, Act: -2.248
R2 dominating r1 over C2. Denote this row as f(r1) ∈ R2
Rollout 1, Act: -2.381
into equations involving t. Let me see.\n\nDenote t = 2y. Then y = t

Layer 4

GATE_PROJ

Top 16 Positive Activations
Rollout 0, Act: 0.012
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 1, Act: 0.012
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 2, Act: 0.012
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 3, Act: 0.012
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 4, Act: 0.012
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 11, Act: 0.012
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 6, Act: 0.012
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 7, Act: 0.012
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 8, Act: 0.012
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 9, Act: 0.012
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 10, Act: 0.012
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 16, Act: 0.012
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 12, Act: 0.012
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 13, Act: 0.012
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 14, Act: 0.012
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 15, Act: 0.012
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Top 16 Negative Activations
Rollout 21, Act: -6.294
→Z with such regularity (like satisfying additivity and certain functional equations) might have to be constant
Rollout 49, Act: -6.295
has countable spread and is T5 (hereditary normal), which we might or might not have,
Rollout 46, Act: -6.298
with angular velocity ��. The acceleration in the inertial frame is a_inertial = a_rot
Rollout 21, Act: -6.300
the image is the entire set Z. So surjective onto Z. So functions in X are surjective
Rollout 46, Act: -6.307
, which are usually part of the potential in an inertial frame. The last term is similar to a
Rollout 81, Act: -6.333
confusing; angle BAC is just one angle at vertex A, but perhaps in terms of coordinate geometry,
Rollout 21, Act: -6.335
1}, which are subsets of Z, not surjective onto Z. So S would not be a subset
Rollout 21, Act: -6.336
holds for the identity function is trivial. For surjective functions, injective functions, since 2f
Rollout 72, Act: -6.338
:\n\nIn a regular pentagon, the diagonal over side is the golden ratio φ = (1 + sqrt
Rollout 1, Act: -6.345
Compare to the above line:\n\nFrom the sum of norms, we have:\n\n-⟨x + x',
Rollout 21, Act: -6.372
. But to have image exactly Z, the surjective. If a function f satisfies f(x -f
Rollout 46, Act: -6.417
to the gradient of U? But only in an inertial frame would the acceleration equal the gradient of the
Rollout 46, Act: -6.431
? However, that would be the case in an inertial frame, but the presence of the centrifugal
Rollout 46, Act: -6.432
\n\nBut these would correspond to the acceleration in an inertial frame minus the centrifugal acceleration? I'm
Rollout 46, Act: -6.495
, without considering the rotating frame, so assuming an inertial frame where U is the conventional gravitational potential,
Rollout 84, Act: -6.614
and "twice as much" are both adverbial phrases modifying the verbs "purchased" and

UP_PROJ

Top 16 Positive Activations
Rollout 7, Act: 3.364
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 44, Act: 3.314
mathrm{T}_{2}$\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to figure out which
Rollout 26, Act: 3.305
mol}^{-1}$ ]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's try to figure this problem
Rollout 98, Act: 3.256
\n(D) 2\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to figure out how
Rollout 6, Act: 3.238
$ . [/*]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through these
Rollout 94, Act: 3.231
\n(D) a,b\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to determine which of
Rollout 18, Act: 3.215
=H_{xy}$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's tackle this problem step by
Rollout 45, Act: 3.193
of $ m$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem where a
Rollout 36, Act: 3.189
. Find $AP.$\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure out how
Rollout 29, Act: 3.182
\rho gH$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure out this
Rollout 23, Act: 3.180
555 Bq\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this problem step by
Rollout 79, Act: 3.179
facility (8,4)\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this crossword clue.
Rollout 54, Act: 3.174
, B = 2\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this question step by
Rollout 1, Act: 3.172
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 11, Act: 3.169
.\n\n*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here where
Rollout 55, Act: 3.157
018}.$\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this matrix A and
Top 16 Negative Activations
Rollout 22, Act: -2.270
T/m)\n\nv_mean = sqrt(8kT/(πm))\n\nAlright, so first I need
Rollout 54, Act: -2.275
ide (strong base)\n\n5. Ethanethiolate (weaker base than hydroxide)\n\n3
Rollout 15, Act: -2.280
- The system can be represented by a tridiagonal matrix \( M \) where each row has
Rollout 54, Act: -2.291
is most hydrated)\n- RS− (thiolate) is a better nucleophile than RO− (
Rollout 54, Act: -2.291
, methanol (neutral), ethanethiolate. Propionate is a carboxylate,
Rollout 54, Act: -2.294
Ka around 10, ethanethiolate (R-S−) would have a base strength
Rollout 54, Act: -2.302
base (strongest acid), ethanethiolate next (still weaker base than hydroxide and
Rollout 26, Act: -2.307
Delta H_f^\circ (\mathrm{HgO}) = \frac{-180.7
Rollout 97, Act: -2.315
may not be precise. Because the reaction stoichiometry is moles, but in terms of partial pressures
Rollout 54, Act: -2.340
bases.\n\nThus, even though ethanethiolate (5) is a weaker base than hydrox
Rollout 54, Act: -2.345
) > hydroxide > ethanethiolate > propionate (weakest base).\n\nThen,
Rollout 54, Act: -2.345
:\n\nOrder: 5 (ethanethiolate) > 2 (hydroxide) >
Rollout 54, Act: -2.410
solution:\n\nIn SN2 reactions, the thiolate ion (e.g., SH−) is a
Rollout 22, Act: -2.426
are:\n\nv_rms = sqrt(3kT/m)\n\nv_mean = sqrt(8kT
Rollout 15, Act: -2.537
Such matrices are called Jacobi matrices or tridiagonal matrices. Nilpotency can be discussed via their
Rollout 15, Act: -2.596
polynomial is x^n.\n\nBut for any tridiagonal matrix, particularly symmetric ones, eigenvalues are real

DOWN_PROJ

Top 16 Positive Activations
Rollout 79, Act: 2.599
use: M, I, N, O, R, C, H, A, N, G
Rollout 79, Act: 2.588
– letters M, I, N, O, R, C, H, A, N, G
Rollout 79, Act: 2.533
changes": M, I, N, O, R, C, H, A, N, G
Rollout 79, Act: 2.529
changes": M, I, N, O, R, C, H, A, N, G
Rollout 79, Act: 2.476
again: M, I, N, O, R, C, H, A, N, G
Rollout 79, Act: 2.465
" : M, I, N, O, R, C, H, A, N, G
Rollout 79, Act: 2.461
changes": M, I, N, O, R, C, H, A, N, G
Rollout 79, Act: 2.446
letters (M, I, N, O, R, C, H, A, N, G
Rollout 79, Act: 2.435
letters: M, I, N, O, R, C, H, A, N, G
Rollout 79, Act: 2.414
letters (M, I, N, O, R, C, H, A, N, G
Rollout 79, Act: 2.409
letters: M, I, N, O, R, H, N, G. Let's see
Rollout 79, Act: 2.395
check: M, I, N, O, R plus H, G, N. That could be
Rollout 79, Act: 2.385
has letters M, I, N, O, R, C, H, A, N, G
Rollout 79, Act: 2.375
letters: M, I, N, O, R, C, H, A, N, G
Rollout 79, Act: 2.373
have are M, I, N, O, R, C, H, A, N, G
Rollout 79, Act: 2.336
) are M, I, N, O, R, C, H, A, N, G
Top 16 Negative Activations
Rollout 79, Act: -1.375
".\n\nSo maybe "minor changes" is an anagram for the answer. Let's try to rearrange
Rollout 49, Act: -1.380
union of closed sets) is called a "perfectly normal" space? Not exactly. Perfectly normal
Rollout 79, Act: -1.382
answer is "something something" that's an anagram of "minor changes" plus some letters. Wait
Rollout 79, Act: -1.382
. So maybe "minor changes" is the anagram fodder, which would form an 8-letter word
Rollout 79, Act: -1.388
". The wordplay usually involves some kind of anagram, reversal, hidden word, etc. "Not
Rollout 39, Act: -1.390
9... part becomes 111... and then 7101 becomes 789
Rollout 79, Act: -1.397
again.\n\nIf "Noticed" is the anagram indicator, then anagram of "minor changes in
Rollout 79, Act: -1.403
's see. If "noticed" is the anagram indicator, then "minor changes" has to be
Rollout 51, Act: -1.424
variable to y = x - ��, and then considering the integral from -θ to 2π
Rollout 79, Act: -1.427
, then "minor changes" has to be anagrammed. Let's check the letters. "Minor
Rollout 74, Act: -1.456
fore-and-aft component K_b (proportional to the Earth's field?) and athwartships
Rollout 57, Act: -1.462
pen sets. Wait, perhaps regularity is a result of the density topology being a regular topology through its
Rollout 79, Act: -1.483
etc. "Noticed" might be the anagram indicator. "Minor changes" could be the letters
Rollout 62, Act: -1.532
−m+11.](https://oeis.org/search?q=%22m%5E
Rollout 79, Act: -1.571
words from these letters include:\n\n- "Hernias" plus some others. No. Maybe "Chair
Rollout 23, Act: -1.590
You are a helpful mathematics assistant.\n<|im_start|>user\nImagine a radioactive nuclei X(Z,A) can decay into

Layer 5

GATE_PROJ

Top 16 Positive Activations
Rollout 34, Act: 0.470
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 47, Act: 0.470
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 63, Act: 0.470
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 55, Act: 0.470
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 84, Act: 0.470
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 78, Act: 0.470
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 5, Act: 0.470
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 60, Act: 0.470
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 43, Act: 0.470
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 94, Act: 0.470
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 10, Act: 0.470
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 15, Act: 0.470
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 11, Act: 0.470
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 12, Act: 0.470
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 13, Act: 0.470
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 14, Act: 0.470
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Top 16 Negative Activations
Rollout 6, Act: -6.376
, no, because the machine M doesn't have prior knowledge of α_n. The machine M needs to
Rollout 28, Act: -6.376
O, A, B, G are given with centroid G. The centroid being (a/3,
Rollout 11, Act: -6.379
where d is some variable. But likely overcomplicating.\n\nAlternatively, assume a = d, b
Rollout 66, Act: -6.386
13,85) is pairwise coprime.\n\nTherefore, maybe this is the correct triplet.
Rollout 81, Act: -6.404
degrees。\n\nConflict arises. The angle calculated here with coordinate system setup is约107 degrees,but tan
Rollout 81, Act: -6.418
Alternatively, notice that previous steps generated this equation through coordinate system. In coordinate system approach, h=5
Rollout 21, Act: -6.426
) -x))=0. But perhaps over complicating.\n\nAlternatively, set equations that take f being
Rollout 1, Act: -6.433
vectors could form an orthogonal set with each vector having norm d / sqrt(2}, so that the distance
Rollout 95, Act: -6.434
t from 0 to1, Pk has coordinates (9t, 0). Similarly, point
Rollout 6, Act: -6.448
n + n^k �� m? Unlikely. For example, m is dominated by n^
Rollout 96, Act: -6.450
, and similarly, any point on line M has coordinate (0, δ).\n\nBut because lines L and
Rollout 24, Act: -6.462
this reasoning may not hold since the domination is over columns in C.\n\nWait, maybe another angle. For
Rollout 96, Act: -6.510
is the origin, any point on line L has coordinate (γ, 0), where γ is the
Rollout 79, Act: -6.525
those letters make an 8-letter word? Unlikely. Let's see. Maybe "Scanning"?
Rollout 81, Act: -6.537
approach we calculated the angle between AB and AC with coordinate system where point A is at (0, h
Rollout 38, Act: -6.613
x is the distance from southeast corner, which has coordinate (0, W). So (-x, W

UP_PROJ

Top 16 Positive Activations
Rollout 69, Act: 1.960
12→a= -4. But a must be a positive integer, so contradiction.\n\nWhich suggests
Rollout 12, Act: 1.939
period depending on a and b. Wait, more precisely, for the integral in the form:\n\n∫_
Rollout 65, Act: 1.900
multiple of 1995. But a must less than or equal to 1995
Rollout 24, Act: 1.857
terms of dominance on C, and columns in C must "cover" all other columns in terms of dominance
Rollout 67, Act: 1.832
= -15d/2.\n\nBut a must divide into 1440 as per equation
Rollout 57, Act: 1.791
y �� U). \n\nWait, but more precisely, closed sets in the density topology are precisely the
Rollout 83, Act: 1.775
,4: c, a, b. Sum must be 0.\n\nContinuing, since the pattern
Rollout 97, Act: 1.771
(in terms of pressure). Wait, maybe more precisely: Because the decomposition reduces N2O5 and
Rollout 29, Act: 1.758
't both be in the options, perhaps the answer is E.\n\nWait, but given that P = rho
Rollout 65, Act: 1.756
*5*7*19. So a must be divisible by 3,5,7,
Rollout 1, Act: 1.755
where each pair is separated by d, perhaps S must lie on such a quadratic manifold. But unless I
Rollout 97, Act: 1.740
have to be careful.\n\nWait, perhaps the answer is to model the partial pressure of N2O5
Rollout 65, Act: 1.736
19=1995. So a must be a multiple of 1995.
Rollout 81, Act: 1.731
I need to recall the formula. Wait, more precisely, in a general triangle, tan A can be
Rollout 46, Act: 1.727
's not strictly "in terms of U". I must misunderstand something. Alternatively, maybe it's asking to
Rollout 65, Act: 1.725
/k. Since c must be integer, and a must be integer: a =k*199
Top 16 Negative Activations
Rollout 87, Act: -2.319
and any necessary numerical constants.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the electric
Rollout 74, Act: -2.323
, and $\theta$.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to derive an expression
Rollout 16, Act: -2.323
by Yannick Yao*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Emilia is trying to grow
Rollout 10, Act: -2.327
matter how his opponent plays.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that in
Rollout 8, Act: -2.337
$m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to
Rollout 58, Act: -2.341
by Yannick Yao*\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. So Patchou
Rollout 93, Act: -2.360
2014}$.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here:
Rollout 79, Act: -2.371
facility (8,4)\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this crossword clue.
Rollout 62, Act: -2.371
four not necessarily distinct primes.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this problem: Find
Rollout 98, Act: -2.374
\n(D) 2\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to figure out how
Rollout 69, Act: -2.384
Team Round #10*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find positive integers
Rollout 72, Act: -2.393
$ \n1. $ \n<|im_start|>assistant\n<|im_start|>think\nAlright, let's tackle this problem: Find
Rollout 42, Act: -2.403
oct-5-ene\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to figure out the
Rollout 83, Act: -2.403
me try to tackle this problem. It's about labeling the vertices of a regular nonagon (9-sided
Rollout 42, Act: -2.694
Answer Choices:\n(A) [1,1'-bi(cyclobutan)]-1-ene\n
Rollout 42, Act: -2.776
options are:\n\n(A) [1,1'-bi(cyclobutan)]-1-ene –

DOWN_PROJ

Top 16 Positive Activations
Rollout 58, Act: 2.957
\nOkay, let's see. So Patchouli has an exam with k parts, right? Each
Rollout 58, Act: 2.826
a helpful mathematics assistant.\n<|im_start|>user\nPatchouli is taking an exam with $k >
Rollout 37, Act: 2.598
could somehow see that each term in the Leibniz formula for determinant involves some commutators,
Rollout 34, Act: 2.473
inside the integrand, we can use Leibniz's rule. The general formula is:\n\nd
Rollout 44, Act: 2.461
is simultaneously both isothermal as well as adiabatic\n\n[C] The work done by the gas
Rollout 12, Act: 2.458
use differentiation under the integral sign, probably Leibniz's rule. \n\nLeibniz's
Rollout 46, Act: 2.445
acceleration are both included in U, and then all that's left is the Coriolis term. Hmm
Rollout 15, Act: 2.416
Because whether the count is odd or even is all that matters.\n\nWait: exactly:\n\nActually, to get
Rollout 42, Act: 2.415
the carbocation undergoes a Wagner-Meerwein rearrangement. If the initial carbocation is at
Rollout 44, Act: 2.373
but if instead the process is reversible and adiabatic, perhaps the work is different.\n\nWait,
Rollout 34, Act: 2.343
, ds \right)\n\]\n\nUsing Leibniz's rule, we get:\n\n\[\nx
Rollout 44, Act: 2.303
V2). Or, if it's adiabatic, different.\n\nBut the question is comparing the
Rollout 3, Act: 2.296
focus might specifically align with one temperature.\n\nGiven all that, but as per my knowledge cutoff is 2
Rollout 44, Act: 2.296
temperature might change, if it's done adiabatically, it's a different calculation.\n\nAlternatively,
Rollout 44, Act: 2.279
othermal, then less. If it's adiabatic, more?\n\nWait, for example, let
Rollout 17, Act: 2.240
0^-7 T·m/A.\n\nGiven all that, let's use calculator steps:\n\nCompute numerator:
Top 16 Negative Activations
Rollout 32, Act: -4.342
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 35, Act: -4.342
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 40, Act: -4.342
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 59, Act: -4.342
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 58, Act: -4.342
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 50, Act: -4.342
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 34, Act: -4.345
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 55, Act: -4.345
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 47, Act: -4.345
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 5, Act: -4.345
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 84, Act: -4.345
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 43, Act: -4.345
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 63, Act: -4.345
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 60, Act: -4.345
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 78, Act: -4.345
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 94, Act: -4.345
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user

Layer 6

GATE_PROJ

Top 16 Positive Activations
Top 16 Negative Activations
Rollout 29, Act: -6.388
answer A.\n\nAnother analogy: if you have a container with two compartments, one open at the top with
Rollout 21, Act: -6.394
) - y =0. Then similarly, that forces f(z) = z for any z that can
Rollout 46, Act: -6.422
rotating frame.\n\nAlternatively, maybe they require using the chain rule in some way? Perhaps we need to take
Rollout 93, Act: -6.432
/32≈1.03 which matches manual. Earlier when testing m=4: sum via
Rollout 71, Act: -6.434
8. a and b as negative integers. Pairs: (-1,-198), (-2
Rollout 93, Act: -6.436
=1.125, which matches our manual computation (9/8). So the formula holds
Rollout 46, Act: -6.453
. Alternatively, maybe it's asking to apply the chain rule via U.\n\nAlternatively, if the energies are
Rollout 62, Act: -6.484
product of four primes (repeats allowed).\n\nBut manual trial may be needed. Let's assume that m
Rollout 54, Act: -6.502
.\n\nWait, no: Let's clarify:\n\nAcid strength: stronger acid has lower pKa. Ethan
Rollout 10, Act: -6.534
in losing positions beyond what I calculated, but Given manual Grundy number calculation up to n=30
Rollout 38, Act: -6.542
to the southeast corner (x).\n\n1. **Travel Time Equations**:\n - Ana's path
Rollout 60, Act: -6.552
**\n\boxed{6}\n\n<|im_start|>answer\nA stationary source emits sound of frequency \( f_0 =
Rollout 93, Act: -6.618
1,m=2 formula gives1,matches. Manual computation matches. For m=3 formula gives3
Rollout 62, Act: -6.626
other factor). No.Maybe else.\n\nOk, after manual's tedious efforts but maybe no other way, to
Rollout 20, Act: -6.682
per earlier.\n\nBut alternatively, let me check with manual calculation for n=LCM(42,
Rollout 33, Act: -6.735
:\n\nWith f(4)=13.\n\nBut manual count is time-consuming. However using inclusion-exclusion

UP_PROJ

Top 16 Positive Activations
Rollout 7, Act: 2.168
$47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve this
Rollout 79, Act: 2.148
facility (8,4)\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this crossword clue.
Rollout 14, Act: 2.090
dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this inequality
Rollout 93, Act: 2.073
2014}$.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here:
Rollout 43, Act: 2.069
0\right\rangle\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure out this
Rollout 26, Act: 2.054
mol}^{-1}$ ]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's try to figure this problem
Rollout 41, Act: 2.053
at most two different digits?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so the problem is asking: How
Rollout 55, Act: 2.052
018}.$\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this matrix A and
Rollout 94, Act: 2.048
\n(D) a,b\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to determine which of
Rollout 3, Act: 2.043
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 63, Act: 2.031
for the three-variable case.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find a function
Rollout 48, Act: 2.030
Does the header contains error?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this IPv4 header
Rollout 30, Act: 2.022
3}}\right)$.\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I've got this problem here
Rollout 99, Act: 2.021
value of x_3?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through this
Rollout 45, Act: 2.011
of $ m$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem where a
Rollout 88, Act: 2.009
channels into a single vector)\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to figure out the
Top 16 Negative Activations
Rollout 42, Act: -2.594
and hexahydro would mean it's partially hydrogenated. Not exactly sure.\n\n(D) spiro[
Rollout 73, Act: -2.621
k = x(1 - (n+1)x^n + nx^{n+1})/(1
Rollout 49, Act: -2.802
Wait, but in the Cantor-Bendixson decomposition, X can be written as the union of
Rollout 49, Act: -2.818
son theorem?\n\nIn the Cantor-Bendixson theorem, we can decompose a Polish space into
Rollout 49, Act: -2.834
ixson derivative.\n\nThe Cantor-Bendixson theorem states that in a separable metrizable
Rollout 49, Act: -2.862
Alternatively, think of the Cantor-Bendixson theorem. If X is a Hausdorff
Rollout 49, Act: -2.886
an approach similar to the Cantor-Bendixson theorem?\n\nIn the Cantor-Bendixson
Rollout 49, Act: -2.913
, we might consider the Cantor-Bendixson derivative.\n\nThe Cantor-Bendixson theorem
Rollout 49, Act: -2.929
in such a space the Cantor-Bendixson derivative.\n\nIn such a space, consider X,
Rollout 49, Act: -2.967
can consider some kind of Cantor-Bendixson decomposition.\n\nAlternatively, since every subset is Fσ
Rollout 37, Act: -2.976
will divide the determinant of an (n-1)x(n-1) matrice with similar structure.\n\n
Rollout 49, Act: -2.979
be built up by its Cantor-Bendixson sequence. At each stage, you remove isolated points
Rollout 49, Act: -2.990
and every level of the Cantor-Bendixson decomposition is countable. But I am not sure
Rollout 49, Act: -2.994
, each level of the Cantor-Bendixson decomposition is a discrete subspace. For Lindel
Rollout 49, Act: -3.043
able. Then apply the Cantor-Bendixson process: removing isolated points transfinitely until we get
Rollout 49, Act: -3.088
topological spaces, the Cantor-Bendixson process can still be applied. Let’s attempt it

DOWN_PROJ

Top 16 Positive Activations
Rollout 0, Act: 6.269
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 1, Act: 6.269
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 6, Act: 6.269
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 3, Act: 6.269
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 4, Act: 6.269
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 7, Act: 6.269
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 14, Act: 6.269
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 9, Act: 6.269
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 15, Act: 6.269
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 12, Act: 6.269
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 10, Act: 6.269
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 11, Act: 6.269
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 8, Act: 6.269
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 16, Act: 6.269
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 17, Act: 6.269
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 20, Act: 6.269
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Top 16 Negative Activations
Rollout 42, Act: -1.492
initial carbocation is at the central carbon (tertiary maybe?), perhaps adjacent to a three-membered
Rollout 79, Act: -1.512
M, A, N, A, G, E, R (but many duplicates). Original letters have
Rollout 54, Act: -1.545
, a bulky nucleophile (like tert-butoxide) is less nucleophilic than hydrox
Rollout 37, Act: -1.629
ant, but the determinant is a specific immanant with characters=sign. But unless there's a
Rollout 37, Act: -1.670
.\n\nAlternatively, this is akin to an immanant, but the determinant is a specific immanant
Rollout 62, Act: -1.732
a product of four primes is called a sphenic number if it's square-free with three distinct primes
Rollout 49, Act: -1.826
not σ-compact. Therefore, the contrapositive would be: If X is uncountable and
Rollout 49, Act: -1.885
, checking reference: In General Topology by Engelking, Theorem 3.8.2 on
Rollout 49, Act: -2.060
is not σ-compact. Therefore, the contrapositive would be: If X is uncountable
Rollout 12, Act: -2.082
\nu<1$ . [P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so
Rollout 49, Act: -2.082
able.\n\nRecall a theorem from [Engelking, General Topology], maybe: A Hausd
Rollout 39, Act: -2.541
999 + \cdots + \underbrace{99\ldots 99}_
Rollout 2, Act: -2.555
9 \times \cdots \times \underbrace{99\cdots9}_{\text
Rollout 39, Act: -2.603
999 + \cdots + \underbrace{99\ldots 99}_
Rollout 2, Act: -2.689
9 \times \cdots \times \underbrace{99\cdots9}_{\text
Rollout 24, Act: -3.492
other, we get same sizes. Maybe use Dilworth's theorem or similar.\n\nAlternatively, think in terms

Layer 7

GATE_PROJ

Top 16 Positive Activations
Rollout 23, Act: 6.827
5 Bq. So someone who miscalculates but also use less approximate e exponent or had a
Rollout 98, Act: 6.610
. The RGB phase is part of post-main-sequence evolution. So if the system is old enough to
Rollout 99, Act: 6.607
if they said 2 iterations, but maybe mistyped. Or, maybe my approach is wrong?\n\nWait
Rollout 43, Act: 6.564
my initial Final Answer line is perhaps I just mistyped when I wrote "\boxed{A}". So according
Rollout 4, Act: 6.532
0.88%, or maybe the answer is exact at 0.87%.\n\nBut wait,
Rollout 98, Act: 6.504
Msun. The K3V is a main sequence star, maybe around 0.7-0
Rollout 98, Act: 6.497
), the A0V is still on the main sequence. So after 80 Myr, A
Rollout 26, Act: 6.495
03.4 is three sig figs at most, so:\n\n-303.4 (
Rollout 98, Act: 6.491
the A0V would still be on the main sequence (since its lifetime is ~370 million
Rollout 99, Act: 6.486
iteration. Alternatively, it's a possible mistranslation/typo, but based on the exact steps
Rollout 98, Act: 6.476
). If the A star really is on the main sequence with 0.9Msun, it can
Rollout 81, Act: 6.474
the angle being measured? \n\nMaybe miscalculating the angle.\n\nAlternatively, recalling that both approaches use
Rollout 98, Act: 6.471
V (lower mass) is still on the main sequence. So, their current state would have the WD
Rollout 98, Act: 6.450
M4V stars to still be on the main sequence.\n\n2. **WD (B5 when MS
Rollout 98, Act: 6.439
The A0V would still be on the main sequence by the time the B5V becomes a WD
Rollout 98, Act: 6.434
V (1Msun) is still on main sequence (lifetime ~10 billion years). The M
Top 16 Negative Activations

UP_PROJ

Top 16 Positive Activations
Rollout 0, Act: 2.461
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 1, Act: 2.461
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 2, Act: 2.461
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 4, Act: 2.461
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 12, Act: 2.461
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 3, Act: 2.461
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 6, Act: 2.461
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 8, Act: 2.461
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 11, Act: 2.461
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 13, Act: 2.461
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 15, Act: 2.461
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 7, Act: 2.461
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 10, Act: 2.461
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 9, Act: 2.461
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 16, Act: 2.461
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Rollout 14, Act: 2.461
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\n
Top 16 Negative Activations
Rollout 93, Act: -2.246
(2014,1007}}。\n\nBut perhaps there's an identity that that expression
Rollout 79, Act: -2.247
Mags". No.\n\nWait, maybe "Organ Chimes". C, H, I, M,
Rollout 79, Act: -2.255
H, O, M, I, N, G – 6 letters. Then using letters: M
Rollout 95, Act: -2.256
在特定时刻, trapezoids会相遇.\n\nHowever,the key lies in the question stating each region
Rollout 95, Act: -2.264
5*2=7350,右.\n\n So Substitute:\n\n= [ (2*36
Rollout 95, Act: -2.273
而得到不同的每个梯形的面积相待,作者可能误解了问题可能允许每个区域面积
Rollout 79, Act: -2.276
M, I, N, O, R, C, H, A, N, G, E
Rollout 79, Act: -2.288
M, I, N, O, R, C, H, A, N, G, E
Rollout 79, Act: -2.290
M, I, N, O, R, C, H, A, N, G, E
Rollout 79, Act: -2.303
, S, R, N. Maybe "Carn SG EN" – no. Wait, letters remaining
Rollout 79, Act: -2.318
M, I, N, O, R, C, H, A, N, G, E
Rollout 79, Act: -2.339
M, I, N, O, R, C, H, A, N, G, E
Rollout 79, Act: -2.356
M, I, N, O, R, C, H, A, N, G, E
Rollout 79, Act: -2.379
M, I, N, O, R, C, H, A, N, G, E
Rollout 79, Act: -2.490
H, O, M, I, N, G (homing). Then remaining letters: C,
Rollout 79, Act: -2.550
"minor" is an anagram of "morin" but that's part of it. Wait,

DOWN_PROJ

Top 16 Positive Activations
Rollout 0, Act: 2.481
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 3, Act: 2.481
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 1, Act: 2.481
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 8, Act: 2.481
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 9, Act: 2.481
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 4, Act: 2.481
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 7, Act: 2.481
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 11, Act: 2.481
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 17, Act: 2.481
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 16, Act: 2.481
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 14, Act: 2.481
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 6, Act: 2.481
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 10, Act: 2.481
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 15, Act: 2.481
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 12, Act: 2.481
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 20, Act: 2.481
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Top 16 Negative Activations
Rollout 25, Act: -2.027
0 \mathrm{~m} / \mathrm{s}$. Although BU had a very small mass estimated
Rollout 25, Act: -2.039
�� 6.8e-5 m/s2. Even such a massive asteroid (1e
Rollout 54, Act: -2.047
This aligns with scenarios where the rate-determining step (RDS) involves a charged reactant
Rollout 60, Act: -2.054
is the speed of the car, 2 m/s. So v_source = 2 m/s\n\n
Rollout 3, Act: -2.059
\(169 \, \text{m/s}\).\n\n**Final Answer:**\n\[\n\
Rollout 3, Act: -2.071
three sig figs justify 169 m/s.\n\nTherefore the valid boxed answer is \boxed{
Rollout 22, Act: -2.072
694} \, \text{m/s}\n\]<|im_end|>
Rollout 22, Act: -2.097
1694 \, \text{m/s} \)\n\n**Final Answer**:\n\[\n
Rollout 27, Act: -2.105
0^{5} \, \text{m/s}\n \]\n\n2. **Estimate the
Rollout 60, Act: -2.113
= 330 \, \text{ms}^{-1} \) is the speed of
Rollout 60, Act: -2.162
approaching the source at a speed of 2 m/s. The sound reflects off the car and comes back
Rollout 25, Act: -2.180
9300 \, \text{m/s} \) and the impact parameter \( r \
Rollout 3, Act: -2.193
\boxed{169 \text{ m/s}}.\n\nHowever, depending on source of problem,
Rollout 60, Act: -2.196
air is $330 \mathrm{~ms}^{-1}$ and the car reflects the sound
Rollout 60, Act: -2.417
speed of \( 2 \, \text{ms}^{-1} \). We need to find
Rollout 60, Act: -2.554
with a speed of $2 \mathrm{~ms}^{-1}$. The reflected signal is received

Layer 8

GATE_PROJ

Top 16 Positive Activations
Rollout 97, Act: 1.375
$\mathrm{Y}$?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I have this
Rollout 54, Act: 1.354
, B = 2\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this question step by
Rollout 42, Act: 1.346
oct-5-ene\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to figure out the
Rollout 26, Act: 1.343
mol}^{-1}$ ]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's try to figure this problem
Rollout 3, Act: 1.308
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 22, Act: 1.305
2) in m/s\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to calculate the mean
Rollout 88, Act: 1.276
channels into a single vector)\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to figure out the
Rollout 42, Act: 1.252
,5,6-hexahydropentalene – Pentalene is a bicyclic compound
Rollout 17, Act: 1.216
$ (in SI units).\n<|im_start|>assistant\n<|im_start|>think\nAlright, let me try to tackle this problem
Rollout 44, Act: 1.203
mathrm{T}_{2}$\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to figure out which
Rollout 42, Act: 1.197
,5,6-hexahydropentalene\n(D) spiro[3.4]
Rollout 42, Act: 1.195
4,5,6-hexahydropentalene\n(D) spiro[3.4
Rollout 43, Act: 1.189
0\right\rangle\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure out this
Rollout 27, Act: 1.185
along the line of sight).\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I've got this problem here
Rollout 42, Act: 1.161
4,5,6-hexahydropentalene – Pentalene is a bicyclic
Rollout 23, Act: 1.146
555 Bq\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this problem step by
Top 16 Negative Activations
Rollout 79, Act: -3.604
team". That's not in the letters. "Unit" – U not present. "Base": B
Rollout 79, Act: -3.607
as the fodder. Let's count: M I N O R C H A N G E S I
Rollout 79, Act: -3.651
changes" and defining investigators' facility is "SCANNING ROOM." Scans + ROOM checks out)*
Rollout 79, Act: -3.686
solution likely is:\n\n**Answer** \n**SCANNING SHOP** \n\nBut honestly, the correct answer
Rollout 79, Act: -3.757
The proper answer confirmed by crossword databases is **SCANNING ROOM**. However, the accurate answer given
Rollout 79, Act: -3.808
the final answer:\n\n**Answer:** \n**SCANNING ROOM** (8,4) \n*(
Rollout 79, Act: -3.824
)*\n\n**Final Answer:** \n**SCANNING ROOM**\n\nHowever, corrections needed after verification. Actual
Rollout 79, Act: -3.832
"monirch saeng" into "SCANNING HOME" — but not fitting.) The proper
Rollout 79, Act: -3.910
letters, so no. Wait, perhaps "Scanning". Contains T? No. Not possible.\n\nAlternatively
Rollout 79, Act: -3.971
" and defining investigators' facility is "SCANNING ROOM." Scans + ROOM checks out)*\n\n
Rollout 79, Act: -3.974
is:*\n\n**Answer:** \n**SCANNING ROOM**\n\n*(Admitted realization: After considerable time
Rollout 79, Act: -3.978
likely is:\n\n**Answer** \n**SCANNING SHOP** \n\nBut honestly, the correct answer all
Rollout 79, Act: -4.001
final answer:\n\n**Answer:** \n**SCANNING ROOM** (8,4) \n*(Letters
Rollout 79, Act: -4.005
Unlikely. Let's see. Maybe "Scanning"? But S is there once, C is there
Rollout 79, Act: -4.076
monirch saeng" into "SCANNING HOME" — but not fitting.) The proper answer
Rollout 79, Act: -4.133
proper answer confirmed by crossword databases is **SCANNING ROOM**. However, the accurate answer given "

UP_PROJ

Top 16 Positive Activations
Rollout 3, Act: 3.247
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 22, Act: 3.239
2) in m/s\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to calculate the mean
Rollout 98, Act: 3.217
\n(D) 2\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to figure out how
Rollout 94, Act: 3.095
\n(D) a,b\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to determine which of
Rollout 17, Act: 3.089
$ (in SI units).\n<|im_start|>assistant\n<|im_start|>think\nAlright, let me try to tackle this problem
Rollout 23, Act: 3.054
555 Bq\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this problem step by
Rollout 80, Act: 3.039
value of $N$ ?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure out this
Rollout 43, Act: 3.017
0\right\rangle\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure out this
Rollout 87, Act: 3.006
and any necessary numerical constants.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the electric
Rollout 60, Act: 3.001
the frequency it has received).\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure out this
Rollout 74, Act: 2.997
, and $\theta$.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to derive an expression
Rollout 82, Act: 2.996
8/69)$.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find an optimal
Rollout 27, Act: 2.981
along the line of sight).\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I've got this problem here
Rollout 44, Act: 2.977
mathrm{T}_{2}$\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to figure out which
Rollout 88, Act: 2.966
channels into a single vector)\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to figure out the
Rollout 46, Act: 2.955
^{2}\right)$.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to find the second
Top 16 Negative Activations
Rollout 33, Act: -1.206
1 ≤ k ≤5. So the desired count is total permutations minus the number of permutations in the
Rollout 13, Act: -1.210
37789*1.03≈1.34391637
Rollout 8, Act: -1.213
240=31/126≈0.246.\n\nThus, the answer
Rollout 13, Act: -1.213
41492*1.03≈1.55796741
Rollout 13, Act: -1.216
72322*1.03≈1.51258972
Rollout 99, Act: -1.216
/1.6054934≈1.1573\n\nThus, x
Rollout 90, Act: -1.220
018 × 0.54 �� 1090. 201
Rollout 31, Act: -1.229
=(1 - ��5)/2~(approx -0.618).\n\nSo perhaps for
Rollout 16, Act: -1.235
06 -1.01587≈4.0347.\n\nWhich as a
Rollout 64, Act: -1.244
2^{2009} =8 �� but we need exact value.\n\nUsing the formula for
Rollout 99, Act: -1.255
66/1.60549≈-2.1531.\n\nThen,
Rollout 89, Act: -1.255
00/3)=200/3≈66.666...\n\nBut, what
Rollout 13, Act: -1.256
18242*1.03≈≈1.3047731
Rollout 70, Act: -1.265
ln(0.2857) �� -1.2528\n\nln(
Rollout 69, Act: -1.268
a+b$ .\n\n\n*2021 CCA Math Bonanza Team Round #10*\n
Rollout 16, Act: -1.316
286 +6.1935≈5.0506.\n\n5.0

DOWN_PROJ

Top 16 Positive Activations
Rollout 15, Act: 2.349
", which is the same as parity (sum neigbors mod2 =1). Therefore, the
Rollout 42, Act: 2.160
a spiro system, the central carbon (spiro atom) is part of both rings. So the
Rollout 42, Act: 2.134
lobutane rings via a shared carbon (spiro carbon), then maybe spiro[3.4
Rollout 2, Act: 2.090
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 13, Act: 2.090
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 18, Act: 2.090
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 32, Act: 2.090
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 40, Act: 2.090
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 19, Act: 2.090
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 31, Act: 2.090
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 35, Act: 2.090
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 61, Act: 2.090
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 50, Act: 2.090
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 59, Act: 2.090
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 67, Act: 2.090
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Rollout 58, Act: 2.090
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>
Top 16 Negative Activations
Rollout 43, Act: -4.089
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nThere is a C-NOT gate where
Rollout 55, Act: -4.089
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet\n\[\mathbf{A
Rollout 47, Act: -4.089
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA train with cross-sectional area $S
Rollout 60, Act: -4.089
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA stationary source emits sound of frequency $
Rollout 63, Act: -4.089
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $P$ denote the
Rollout 19, Act: -4.096
$ , the remainder is $17$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have
Rollout 21, Act: -4.110
}\n\n(b) \boxed{\text{Yes}}\n\n<|im_start|>answer\n(a) To show that Carl can choose
Rollout 63, Act: -4.114
, algebra A1 for the three-variable case.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need
Rollout 31, Act: -4.147
. Find all possible values of $y$.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's try
Rollout 45, Act: -4.170
the greatest possible value of $ m$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have
Rollout 57, Act: -4.243
\boxed{\text{Regular but not normal}}\n\n<|im_start|>answer\nTo show that the requirement defines a topology
Rollout 32, Act: -4.308
M}^{2009}.$\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need
Rollout 69, Act: -4.328
Final Answer**\n\boxed{204}\n\n<|im_start|>answer\nGiven the equation:\n\n\[\n\frac
Rollout 87, Act: -4.370
(1 + \sqrt{2}) }.\n\n<|im_start|>answer\nTo find the electric potential at a corner
Rollout 51, Act: -4.518
3} \sin 8\theta}.\n\n<|im_start|>answer\nTo evaluate the integral \(\int_
Rollout 86, Act: -4.628
{\dfrac{\pi}{16}}.\n\n<|im_start|>answer\nTo find the minimum value of the integral

Layer 9

GATE_PROJ

Top 16 Positive Activations
Rollout 38, Act: 4.142
arrive at D at the same time. So, perhaps the distances they travel are different. Wait, that
Rollout 95, Act: 4.104
k/N. Which explains discrepancy in results.\n\nSo perhaps the parametric division of AB and AC into equal
Rollout 24, Act: 4.058
interaction between both row dominating and column dominating.\n\nSo perhaps using the minimality in both row and column directions
Rollout 21, Act: 4.055
, where S only contains identity function. So, perhaps adjust the functional equation so that solution must be functions
Rollout 97, Act: 4.039
contribution is straightforward.\n\nSo the initial assumption in this think block was that theY would be4.6
Rollout 46, Act: 4.032
and position are independent variables in mechanics. So, perhaps the intended answer does not include the Coriolis
Rollout 56, Act: 4.025
But the problem states it’s a triangle, so perhaps A2A3A4 or A1A
Rollout 23, Act: 4.009
per minute, but activity is per second. So perhaps the answer is in per second, so lambda_alpha
Rollout 24, Act: 4.002
different minimal pairs could interchange columns and rows. So perhaps there's symmetry here. However, still, how
Rollout 56, Act: 3.993
6, ��3 /12 )\n\nSo perhaps the new triangle is A2A3A4
Rollout 33, Act: 3.983
if n=1, would not apply. So perhaps not useful.\n\nWait, for n=2,
Rollout 7, Act: 3.982
equations, but led to impossibility. Thus, perhaps the answer is different.\n\nAlternatively, maybe N=
Rollout 24, Act: 3.981
is constant. Similarly, same for C. So perhaps, saddle pairs’ row sets form a matroid
Rollout 27, Act: 3.980
_Be, and hence estimate T_c".\n\nSo perhaps, steps are first compute v_rms from Δ
Rollout 21, Act: 3.979
empty. But we need S nonempty. So perhaps that approach is wrong.\n\nAlternative idea: Suppose that
Rollout 89, Act: 3.971
/5th of the children are girls".\n\nSo perhaps the problem maker mixed up the gender in the fractions
Top 16 Negative Activations
Rollout 24, Act: -0.464
of all objects sharing all attributes in B, and B is the set of all attributes shared by all objects
Rollout 49, Act: -0.464
2^{2^{d(X)}}, where d(X) is the density. But if our space
Rollout 57, Act: -0.469
$ x$ whose density at $ x$ equals $ 1$ . Show
Rollout 48, Act: -0.486
612 4000 4006 6F80 0A
Rollout 1, Act: -0.493
$ y\in\mathcal{H}$ such that \n\[ \left\{\frac
Rollout 1, Act: -0.527
able) in $ \mathcal{H}$ such that the distance between any two distinct points
Rollout 1, Act: -0.547
user\nLet $ \mathcal{H}$ be an infinite-dimensional Hilbert space, let
Rollout 48, Act: -0.560
612 4000 4006 6F80 0A
Rollout 1, Act: -0.580
$ S$ is equal to $ d$ . Show that there is a point $ y
Rollout 1, Act: -0.581
>0$ , and suppose that $ S$ is a set of points (not necessarily count
Rollout 1, Act: -0.611
distance between any two distinct points in $ S$ is equal to $ d$ . Show
Rollout 48, Act: -0.618
6F80 0A00 008B 5BC6 AEE0
Rollout 57, Act: -0.655
Let a neighborhood basis of a point $ x$ of the real line consist of all Lebes
Rollout 57, Act: -0.691
Lebesgue-measurable sets containing $ x$ whose density at $ x$ equals
Rollout 1, Act: -0.747
Hilbert space, let $ d>0$ , and suppose that $ S$ is
Rollout 57, Act: -0.836
$ x$ whose density at $ x$ equals $ 1$ . Show that

UP_PROJ

Top 16 Positive Activations
Rollout 93, Act: 2.105
T_m(x)= ±1.\n\nBut for sum_{j}[x_j^^{2m}], does
Rollout 16, Act: 2.094
the sum as we derived, which is:\n\nSum_{k=1}^\infty [1 -
Rollout 56, Act: 2.091
1/2^{n+1})= Sum_{k=0}^n (1/2
Rollout 37, Act: 2.078
is a homogeonous polynomial of degree sum_{i=1}^n k_i. Then
Rollout 1, Act: 2.077
: Let me define y such that y = sum_{i=1}^n a_i x_i
Rollout 93, Act: 2.055
+2))/(4).\n\nIn general,for sum_{k=1}^m cos^{2m
Rollout 93, Act: 2.048
Is there a generating function or identity known for sum_{k=1}^m cos^{2m
Rollout 37, Act: 2.040
a_i are arbitrary integers, the determinant is Sum_{σ} sgn(σ) ��_i
Rollout 64, Act: 2.033
)\n\n Wait, step by step:\n\nFor eachy >=0 given, such that a_{201
Rollout 49, Act: 2.030
Hausdorff space.\n\nSo X = union_{n=1}^∞ K_n, where
Rollout 64, Act: 2.030
a1= x, then total solutions is Sum_{ y=0 }to8 ) sum_( x
Rollout 16, Act: 2.027
of n geometric(p) variables is equal to sum_{k=1}^{infty} [1
Rollout 16, Act: 2.016
formula:\n\nFor n=1, expectation is sum_{k=1}^\infty [1 -
Rollout 64, Act: 2.008
uting z =x -1, expressions:\n\nSum_{y=0}^8 (M -Ky
Rollout 93, Act: 2.006
) are roots of a certain polynomial, then Sum_{k=1}^{1007}
Rollout 93, Act: 2.003
, and in our approach resulted in the formula sum_{k=1}^m [cos(pik
Top 16 Negative Activations
Rollout 84, Act: -0.999
answer is \(\boxed{37}\).<|im_end|>
Rollout 44, Act: -1.005
\neq \mathrm{T}_{2}$\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need
Rollout 39, Act: -1.009
Final Answer**\n\boxed{342}\n\n<|im_start|>answer\nTo find the sum of the digits of
Rollout 2, Act: -1.016
is \(\boxed{109}\).<|im_end|>
Rollout 0, Act: -1.020
Final Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and
Rollout 18, Act: -1.039
(\boxed{(7, 5)}\).<|im_end|>
Rollout 6, Act: -1.049
) to decide \( x \in L \).<|im_end|>
Rollout 54, Act: -1.050
3 and 4, B = 1<|im_end|>
Rollout 19, Act: -1.059
**Final Answer**\n\boxed{78}\n\n<|im_start|>answer\nTo solve the problem, we need to
Rollout 63, Act: -1.059
\boxed{f(p,q) = pq}\n\n<|im_start|>answer\nTo find the function \( f(p,
Rollout 70, Act: -1.072
is \(\boxed{315}\).<|im_end|>
Rollout 69, Act: -1.078
Final Answer**\n\boxed{204}\n\n<|im_start|>answer\nGiven the equation:\n\n\[\n\frac
Rollout 40, Act: -1.084
**Final Answer**\n\boxed{19}\n\n<|im_start|>answer\nThe particle starts at position \((5
Rollout 92, Act: -1.108
{3.78} diopters.<|im_end|>
Rollout 41, Act: -1.111
is \(\boxed{927}\).<|im_end|>
Rollout 79, Act: -1.128
ANGING ROOMS** (8,4).<|im_end|>

DOWN_PROJ

Top 16 Positive Activations
Rollout 16, Act: 2.308
>= k. Since each Ti is >= k with probability (1/2)^{k - 1
Rollout 85, Act: 2.230
. For example, let's suppose we start alternately labeling.\n\nFrom (0,0)=0,
Rollout 10, Act: 2.222
the pattern is adding 2 then 3 alternately. Hmmm. If this continues, then the
Rollout 16, Act: 2.198
" each remaining coupon (sapling) independently with probability 1/2.\n\nBut is there an analogous
Rollout 63, Act: 2.170
1. Each of these directions is taken with probability 1/2, but since this is a
Rollout 33, Act: 2.163
3.\n\nLet's compute f(3).\n\nAlternately, using inclusion-exclusion:\n\nTotal permutations:6
Rollout 10, Act: 2.153
**Understanding the Game**:\n - Players take turns removing square numbers (1, 4,
Rollout 11, Act: 2.063
zero, but in a non symmetric way.\n\nAlternately, consider all three variables with the same modulus,
Rollout 16, Act: 2.032
into a tree. Each sapling independently succeeds with probability 1/2 each day. So the process
Rollout 1, Act: 2.012
istant set have to be of this form?\n\nAlternately, let me try to create y. Assume that
Rollout 10, Act: 1.995
opponent plays, we analyze the game where players alternately remove a square number of pebbles. The
Rollout 10, Act: 1.982
on the table. Two players make their moves alternately. A move consists of taking off the table
Rollout 10, Act: 1.953
me try to understand the game first.\n\nPlayers take turns removing a square number of pebbles (1
Rollout 59, Act: 1.934
coins and a third coin that comes up heads with probability $\frac47$ . Jackie flips the three
Rollout 42, Act: 1.916
bond such that those two rings are now both connected via a single carbon (spiro) with a double
Rollout 10, Act: 1.906
Hmm.\n\nAlternatively, looking at their parity: alternately even and odd. But maybe every other pair.
Top 16 Negative Activations
Rollout 38, Act: -2.039
=(-x, W). So must have -x >= -L, which impliesx <=L. Since
Rollout 65, Act: -2.046
prisms.\n\n Let’s try for prism likea <=B=1995<=c. When
Rollout 65, Act: -2.048
/A⇒A= C2/B.\n\n Additionallya<=B<=C。Given that, in this case
Rollout 7, Act: -2.050
0). Their new scores are original+5 (<=64). Average should be47. So
Rollout 75, Act: -2.074
collect all the split squares, which would amount to <=3 +3 +3=9 squares. But
Rollout 38, Act: -2.099
because since L should be positive, and x must <=L, since otherwise D would lie outside the field
Rollout 42, Act: -2.106
between the two rings. The oct-5-ene part suggests an eight-carbon chain with a double
Rollout 86, Act: -2.115
/2], x >= sinx since sinx <=x for x >=0, and cosx <=
Rollout 42, Act: -2.125
iro[3.4]oct-5-ene would have a spiro union where one ring is
Rollout 25, Act: -2.146
541 km altitude.\n\nNeed Total Δx >= 3541 km.\n\nThus,\n\n(
Rollout 42, Act: -2.163
iro[3.4]oct-5-ene – A spiro compound with two rings, one
Rollout 38, Act: -2.207
. So x cannot exceed L. Hence,\n\nx <= L\n\nGiven L=155/1
Rollout 42, Act: -2.217
'-bi(cyclobutan)]-1-ene\n(B) 1,2-dimethyl
Rollout 42, Act: -2.259
iro[3.4]oct-5-ene\n<|im_start|>assistant\n<|im_start|>think\nOkay, so
Rollout 38, Act: -2.313
have -x >= -L, which impliesx <=L. Since L=155/1
Rollout 86, Act: -2.330
x with sinx.\n\nWait, but 0 <=x <=pi/2. Here, sinx

Layer 10

GATE_PROJ

Top 16 Positive Activations
Rollout 54, Act: 3.360
Wait, maybe I'm confusing things. Let's recall that in polar aprotic solvents, nucle
Rollout 79, Act: 3.330
3. No.\n\nHmm.\n\nWait, I recall hearing the term "Scene of Crime" – that
Rollout 22, Act: 3.295
K. But maybe check both.\n\nBut let me recall. If I look up the formula for mean speed
Rollout 23, Act: 3.258
But maybe the precise calculation is different. Let me recall that e^-70= approx approx
Rollout 69, Act: 3.247
3). Perhaps using multiple-angle identities.\n\nLet me recall the identity: sin(3 theta) =
Rollout 73, Act: 3.245
code here, maybe simulate.\n\nAlternatively, if I write down exponents and compute terms.\n\nAlternatively, maybe
Rollout 12, Act: 3.237
of power functions. But not sure. Let me recall the Beta function:\n\nB(p, q) =
Rollout 27, Act: 3.236
formula for T_c in BEC.\n\nLet me recall the formula for T_c. For Bose-Einstein
Rollout 86, Act: 3.222
to ��x2 cosx. Let me recall.\n\nWait, let's compute ��x2
Rollout 33, Act: 3.211
=3: -, etc. Wait, let me recall.\n\nIn inclusion-exclusion, the general term for
Rollout 86, Act: 3.205
16. Therefore minimum.\n\nBut wait let me recall other methods. Perhaps apply inequality.\n\nFor example,
Rollout 86, Act: 3.200
��x cos2x dx?\n\nLet me recall. No, but integrate ��x cos2
Rollout 15, Act: 3.200
ent over GF(2).\n\nNow, let me recall: for the adjacency matrix of a path graph,
Rollout 45, Act: 3.193
(0) using the recurrence relation. Let me recall the recurrence. Earlier, when we set x +
Rollout 75, Act: 3.184
is 32 possible? Wait, let's recall that 32 can also be written as sums
Rollout 79, Act: 3.183
is the answer here...\n\nWait, I think I recall that some crosswords use terms like "Safe House
Top 16 Negative Activations
Rollout 1, Act: -0.635
x'||2 = 0 ⇒ ⟨x - x', t⟩ = ||x||2 - ||
Rollout 54, Act: -0.636
so ethanethiolate would be CH3CH2S−. Okay, so all nucleoph
Rollout 1, Act: -0.643
S' is ||z|| = ||x - x0||, but in the original set S,
Rollout 46, Act: -0.665
follows:\n\nρ1 = sqrt( (x - x1)^2 + (y - y1)^
Rollout 1, Act: -0.677
H}$ such that the distance between any two distinct points in $ S$ is equal to
Rollout 46, Act: -0.703
/ ��12 * ( (x - x1)/ρ1, (y - y1
Rollout 1, Act: -0.717
= x0 + z' is ||z - z'|| = d. Now, we need to find
Rollout 54, Act: -0.740
ethiolate. Ethanethiol is CH3CH2SH, so ethanethiolate would
Rollout 46, Act: -0.762
then ��1 = sqrt( (x - x1)^2 + (y - y1)^
Rollout 54, Act: -0.774
, ethanethiolate would be CH3CH2S−)\n\nWait, the names:
Rollout 54, Act: -0.778
anethiolate (HS− or CH2CH2S−), is a weaker base than hydro
Rollout 46, Act: -0.789
then ��1 = sqrt( (x - x1)^2 + (y - y1)^
Rollout 54, Act: -0.795
resonance stabilization?\n\nWait, propionate (CH2CH2COO−) is a resonance-stabil
Rollout 54, Act: -0.840
.88, so propionate (CH2CH2COO−) is a weaker base than
Rollout 54, Act: -0.868
:\n\n5. Ethanethiolate (CH3CH2S−): most polarizable, weaker base
Rollout 54, Act: -0.955
OH−)\n3. Propionate (CH2CH2COO−)\n4. Methanol (

UP_PROJ

Top 16 Positive Activations
Rollout 65, Act: 1.944
be cut so to give a similar.\n\nBut, suppose the prism P(a,1995,c
Rollout 65, Act: 1.917
2: Ratio k if permute.\n\nSuppose:\n\na /1995 =19
Rollout 65, Act: 1.908
R. If ratios new prism order different.\n\nSuppose the new prism is considered ordered: either a <=
Rollout 21, Act: 1.878
So ceil(x) is allowed.\n\nAnother example, suppose we define f(x)= floor(x) if x
Rollout 62, Act: 1.851
but then the multiplicity requires the power.\n\nSuppose for 11*11*11
Rollout 65, Act: 1.851
scaled version. Again the same impasse.\n\nSuppose similarity requires that the new proportions must equal the old
Rollout 28, Act: 1.837
there a better way to express it?\n\nAlternatively, suppose, set parameters or express in terms of centroid's
Rollout 65, Act: 1.834
1: dimensions are scaled by k.\n\nSuppose the similar prism has sides ka, kb, kc
Rollout 65, Act: 1.826
to no solutions.\n\nAlternatively, another permutation?\n\nSuppose after cutting c1, then new prism's ratios
Rollout 93, Act: 1.826
m,m))/4^{m}.\n\nNow, suppose we consider the central term C(2m,m
Rollout 27, Act: 1.818
~1.49e5 m/s. Which confirms its the same result.\n\nHence, given
Rollout 65, Act: 1.813
k*c].\n\nPossibly re-ordered:\n\nSuppose original ordered [a,1995,
Rollout 11, Act: 1.807
π. Forcing abc=1.\n\nHowever, suppose arranging a =k e^{iθ}, b
Rollout 1, Act: 1.787
)/2.\n\nPutting all this together.\n\nNow, suppose that we can model some kind of system.\n\nLet
Rollout 65, Act: 1.782
1995,1995). Which is similar to the original prism (a,1
Rollout 14, Act: 1.782
the function such that multiple variables are involved.\n\nSuppose I make a substitution x= cy where c is
Top 16 Negative Activations
Rollout 49, Act: -1.763
dorff space $X$ is $\sigma$ -compact, then $X
Rollout 45, Act: -1.765
: Z \to Z$ be such that $ f(1) \equal{} 1
Rollout 96, Act: -1.780
Let's compute that first.\n\nLine L has slope 5/12 and passes through point A(
Rollout 76, Act: -1.781
that have at least one 2x2 all-red square, and then subtract that from the total to
Rollout 57, Act: -1.792
sets containing a point \( x \) with density 1 at \( x \).\n\n### Regularity\n\n
Rollout 33, Act: -1.795
need to find the number of permutations of the numbers 1 through 6 such that for each k from
Rollout 56, Act: -1.796
starting with a right-angled triangle, with angles 90°, 60°, and
Rollout 45, Act: -1.796
{}4m$ for all non-negative integers $ x$ , find the greatest possible value of
Rollout 5, Act: -1.798
zoid into two regions with areas in the ratio 2:3. Then, we need to find
Rollout 81, Act: -1.801
A\) dividing \(BC\) into segments of length 3 and 17, we need to find
Rollout 28, Act: -1.825
$G$ be the centroid of the triangle $OAB.$ $(a)$ Pro
Rollout 63, Act: -1.837
P$ denote the set of all ordered pairs $ \left(p,q\right)$ of
Rollout 15, Act: -1.873
lamps are placed in a line. At minute 0, some lamps are on (maybe all of
Rollout 11, Act: -1.889
$c$ are three complex numbers with product $1$ . Assume that none of $
Rollout 14, Act: -1.902
0,\infty)$ is integrable and $$ \int_0^\infty f(x
Rollout 72, Act: -1.983
(not necessarily distinct) roots of unity of order $ n $ whose sum is $ \n

DOWN_PROJ

Top 16 Positive Activations
Rollout 11, Act: 2.277
\overline{z} is real (the modulus squared). Since p is a combination of a,
Rollout 11, Act: 2.175
So they can’t be real, even if their modulus is not 1. So all of a,
Rollout 11, Act: 2.029
c= e^{-i2θ}. But modulus of a and b is 1, which is prohibited
Rollout 87, Act: 1.992
795\n\nOther combinations. For instance, considering hypergeometric functions or similar, there is no
Rollout 84, Act: 1.968
.e., \( \frac{1}{2} \) units) and drank twice as much (i
Rollout 28, Act: 1.961
symmetric structure in a and b. So, maybe considering dividing terms appropriately.\n\nAlternatively, suppose let’s set
Rollout 49, Act: 1.954
able Hausdorff space, then there exists some subspace of X that is not σ-compact
Rollout 1, Act: 1.934
the distance between any two is the same, they form a regular simplex. The regular simplex has the property
Rollout 18, Act: 1.932
generated by two elements (by the structure theorem for finitely generated abelian groups), or even a
Rollout 76, Act: 1.914
) or overlapping in a single square.\n\nTherefore, considering all pairs of 2x2 squares, let
Rollout 11, Act: 1.909
1. Let k be a complex number with modulus not 1, e.g., k =
Rollout 81, Act: 1.905
trust? Let's note:\n\nIn coordinate system, considering angle atA as angle between AB(h/3
Rollout 1, Act: 1.875
But actually no, in the previous, if we take multiple points, we need to ensure that they lie
Rollout 37, Act: 1.871
saw earlier isn't the case. For example, taking all a_i distinct primes.\n\nWait, perhaps we
Rollout 27, Act: 1.871
nuclei, \( V_{\text{Be}} \), and estimate the critical temperature \( T_{\
Rollout 60, Act: 1.869
receives the sound. The observed frequency \( f' \) is given by the Doppler effect formula for
Top 16 Negative Activations
Rollout 12, Act: -1.883
). Start with:\n\nf(ν) = ��_{ν}^1 [ (t2 -
Rollout 51, Act: -1.887
sinx| dx = 2 ��0^{π} cos8x sinx dx =
Rollout 17, Act: -1.888
r/a0} (1/r3) ��0^π (3 cos2θ -
Rollout 51, Act: -1.895
appropriately. So the integral becomes:\n\nIntegral = ��0^θ sin(8x)*(-sin
Rollout 12, Act: -1.895
to1. So f(ν) = ��_ν^1 [1 / sqrt( (
Rollout 51, Act: -1.915
sin8t (-sint) dt = ��0^{π} sin8t sint dt\n\n
Rollout 12, Act: -1.926
is:\n\nF(φ, k) = ��_0^phi [ 1 / sqrt(
Rollout 26, Act: -1.931
0083 * 298 �� 0.0083 * 3
Rollout 9, Act: -1.933
_upper and y_lower from a to b is ��[a to b] (y_upper - y
Rollout 12, Act: -1.953
is:\n\nF(φ, k) = ��_0^φ [ 1 / sqrt(
Rollout 17, Act: -1.973
u2 -1) (-du) = ��_{-1}^1 (3u2
Rollout 86, Act: -1.975
pi/2.\n\nSo now compute 3 ��x2 cos x dx. Integrate by parts
Rollout 17, Act: -1.976
, the integral becomes:\n\n**I** = �� [ (μ0 / 4π) (
Rollout 38, Act: -1.993
/6.2)(W + x) �� (0.3871)(W +
Rollout 87, Act: -2.017
1/2) r2 ln r - �� (1/2) r dr = (1
Rollout 87, Act: -2.105
= (1/(4πε0)) �� (ρ / r) dV, where �

Layer 11

GATE_PROJ

Top 16 Positive Activations
Rollout 90, Act: 4.052
itself is 4 digits). Wait, let me think again.\n\nWait, hold on. 20
Rollout 89, Act: 4.052
:2 boys to girls? Wait, let's think. If the ratio of girls to total is
Rollout 91, Act: 4.047
BD is 45. Hmm. Let's think.\n\nBut perhaps it's better to assign coordinates.
Rollout 96, Act: 4.037
are found by perpendicular projections. Wait, let me think.\n\nSuppose in the new coordinate system:\n\n-
Rollout 95, Act: 3.981
is added incrementally. Wait, maybe let's think of the area of triangle APkQk is
Rollout 36, Act: 3.978
let's also consider Equation (4). Let me think. Let's see Equation (4):\n\nx2
Rollout 28, Act: 3.966
2b2/a2.\n\nHmm. Let me think of this as a quadratic form in variables A and
Rollout 20, Act: 3.959
but need to check carefully. Wait, let's think:\n\nBut for modulus 43: n �
Rollout 24, Act: 3.958
might get too complicated.\n\nAlternative approach. Let's think about if we can show |R1| =
Rollout 79, Act: 3.942
. But sometimes, parts can overlap. Let me think again.\n\nIf "Noticed" is the an
Rollout 8, Act: 3.935
want to cross-verify.\n\nWait, let’s think of another approach. For instance, imagine that we
Rollout 36, Act: 3.927
call this equation (d)\n\nBut let's also think about equation (1). From equation (1):\n\n
Rollout 54, Act: 3.923
ophilic?)\n\nBut this is confusing. Let's think in terms of specific examples.\n\nIn aqueous solution
Rollout 29, Act: 3.912
levels are different because of pressure differences. Let's think of it as a U-tube. A U
Rollout 9, Act: 3.907
to EM, h can be anything. Let's think. If h is less than 10,
Rollout 85, Act: 3.905
determined in a different way.\n\nWait, let's think of the problem as similar to a grid where each
Top 16 Negative Activations
Rollout 95, Act: -2.254
3675 )= (15 sqrt(3))/(2*3675 )=
Rollout 97, Act: -2.256
k t}\n\nThus, solving yields t= ln(10}/(2k))=2.
Rollout 97, Act: -2.260
Y is 2.30, since ln(10)/ (2 k}= ln(1
Rollout 87, Act: -2.263
ε0) [ π/3 - (sqrt(3) - 1)/2 ]\n\nBut I
Rollout 14, Act: -2.275
+ a/2.\n\nNeed to show ( sqrt(2 K a ) +a /2 )3
Rollout 14, Act: -2.287
cube we have:\n\nI^3 <= (sqrt( J) sqrt(A ) + K/A )^
Rollout 95, Act: -2.290
0。\n\n To have this length rational, (sqrt(3) k)/490 must equals rational
Rollout 87, Act: -2.295
)/(4 π ε0) * (3 ln(1 + ��2) + ��3
Rollout 87, Act: -2.337
by step. Therefore, expression is:\n\n3 ln(1 + sqrt(2)) + sqrt(3
Rollout 56, Act: -2.342
A2|= (1/14)(sqrt(559)/12)= sqrt(5
Rollout 14, Act: -2.343
a )(a/2 ) +3 (sqrt(2 K a ))(a2/4 )
Rollout 95, Act: -2.360
) )/3675= (sqrt(675)/2)/3675
Rollout 97, Act: -2.373
if rounded,) but likely the precise answer isln(10}/(2k≈5×1
Rollout 97, Act: -2.387
=1.45→ t= ( ln(10)/ (2 k))=23
Rollout 97, Act: -2.405
.45} leads to t= (ln(10))/(2×5e-4)}
Rollout 95, Act: -2.471
)。\n\n Thus length of each segment is (sqrt(3) k)/490。\n\n To have

UP_PROJ

Top 16 Positive Activations
Rollout 14, Act: 2.537
/ k^3. So it's even then not tight. Therefore, maybe the 8 here
Rollout 62, Act: 2.488
. Not that easy to do manually.\n\nAlternatively, think of m=55:\n\nm=55
Rollout 95, Act: 2.438
50.\n\n But here is the distance. So then, area of trapezoid would be average
Rollout 41, Act: 2.424
a a, etc. Thus 6. So then for each pair, 6 numbers. So
Rollout 79, Act: 2.423
A, G, S).\n\nSo that's also not possible.\n\nHmm. After all this time, I
Rollout 75, Act: 2.413
here 8*4=32. So then need to express32 as sum of three even
Rollout 75, Act: 2.386
1+1+1+1. So then 49 +4 (remaining)+4 +
Rollout 62, Act: 2.381
1) should be product of two primes. So then 11 multiplied by these two primes would make
Rollout 75, Act: 2.376
4 +4 +0= allowing zero. So then scaling back, sqrt(4×4)=4
Rollout 51, Act: 2.376
/63. That part checks out. So then the answer is -4/63 sin8
Rollout 96, Act: 2.375
indicating calculation messed sign or original parameters.\n\nWait, thinking back: When we solve the vector equation: OP
Rollout 75, Act: 2.364
62+32+22. So then total three squares. But if I need three squares
Rollout 23, Act: 2.363
activity, not Total activity or beta decay.\n\nSo then why is D close to the total activity.\n\nAlternatively
Rollout 11, Act: 2.350
Even this direction is not leading anywhere. Finally, think of the answer. Since in olympiad problems
Rollout 75, Act: 2.348
2 + (2a4)^2. So then, total 8M +3 = sum of
Rollout 75, Act: 2.339
4 +4+1=9. So then, 25 + (4 +4 +
Top 16 Negative Activations
Rollout 87, Act: -1.505
me let u = ln(1 + sqrt(1 + r2)), dv = r dr\n\nThen
Rollout 51, Act: -1.512
≤ �� ≤ π, �� + π ≤ 2π when �� ≤ π. Wait
Rollout 14, Act: -1.513
ity to show:\n\n\[\nI^3 \leq 8JK.\n\]\n\nThus, the
Rollout 87, Act: -1.517
�0^1 ln[(1 + sqrt(1 + ��2 + ��2))/sqrt
Rollout 14, Act: -1.533
fty f(x) dx \right)^3 \leq 8\left(\int_0^\
Rollout 18, Act: -1.539
H is the kernel of the homomorphism φ: Z2 → Z/xZ defined by φ(k
Rollout 14, Act: -1.541
then our goal reduces to showing that I^3 ≤8, which is equivalent to I≤2.
Rollout 64, Act: -1.552
, the inequality a_{2010} ≤ 2^{2012} becomes:\n\n
Rollout 14, Act: -1.579
inequality $(a + b + c)^3 \leq ...$. Hmm, but not directly applicable here
Rollout 14, Act: -1.593
1. But then the inequality requires I^3 ≤8J K, which would give 8 as
Rollout 14, Act: -1.596
but the question states to prove that I^3 ≤8 J K. Here, if I≈1
Rollout 14, Act: -1.614
We need to show that \( I^3 \leq 8JK \).\n\nWe split the integral
Rollout 14, Act: -1.614
=t. Then the inequality states that I^3 ≤8 *1 *t.\n\nTo make this work
Rollout 14, Act: -1.627
1. Then need to show that I^3 ≤8 *1 *1→ I ≤2.\n\n
Rollout 14, Act: -1.654
me see.\n\nWe need to relate I^3 ≤8 J K.\n\nBy AM_ GM,\n\nI
Rollout 14, Act: -1.702
We need to show that $I^3 \leq 8 J K$.\n\nHmm. So

DOWN_PROJ

Top 16 Positive Activations
Rollout 25, Act: 3.417
with 3 significant digits. Use $6357 \mathrm{~km}$ as the radius
Rollout 62, Act: 2.553
663. Check. 2663. Likely prime?\n\nSimilarly,maybe it will take
Rollout 81, Act: 2.379
adratic solution: h= [-70 ±sqrt(70^2 +4*11*
Rollout 81, Act: 2.335
Quadratic solution:\n\nh=(70 ±sqrt(4900 +4*11*
Rollout 86, Act: 2.335
pi/4, cosx is greater than sqrt(2)/2≈0.707,
Rollout 7, Act: 2.329
16x =71N -209N/3= (213N -
Rollout 25, Act: 2.326
radius of the Earth is provided as 6357 km. The question specifies that we should consider
Rollout 25, Act: 2.301
surface, which is Earth's radius 6357 km + 3541 km =
Rollout 38, Act: 2.284
quadratic formula:\n\nr=(625±sqrt(6252 -4*168
Rollout 65, Act: 2.270
2/c=19952 /sqrt(1995*a)=1995
Rollout 7, Act: 2.262
x.\n\nTherefore, the equation becomes:\n\n209N/3 +16x =71
Rollout 7, Act: 2.239
3= (213N -209N)/3=4N/3.\n\nTherefore
Rollout 28, Act: 2.221
2. So eigenvalues are (8 ± sqrt(64 - 48))/2 = (
Rollout 25, Act: 2.210
igee equal to Earth's radius (6357 km). Assume the same speed at infinity (
Rollout 87, Act: 2.208
in the first quadrant up to a radius of sqrt(2)*a). Wait, but integrating from y
Rollout 95, Act: 2.204
length P1Q1 would be BC*sqrt(1/2450) => (5
Top 16 Negative Activations
Rollout 26, Act: -2.557
from $298.0 \mathrm{~K}$ to $312.8 \
Rollout 60, Act: -2.562
source with a speed of $2 \mathrm{~ms}^{-1}$. The reflected signal is
Rollout 54, Act: -2.568
ethanethiolate (HS− or CH2CH2S−), is a weaker base than
Rollout 4, Act: -2.613
\pm 0.02 \mathrm{~V}$. What is the percentage error in the
Rollout 82, Act: -2.667
need to modify the Huffman algorithm for a mixed-radix system? Let me recall how Huffman coding works.
Rollout 48, Act: -2.671
hex dump with spaces as given:\n4500 0034 B612
Rollout 44, Act: -2.672
(\mathrm{p}_{1}, \mathrm{~V}_{1}, \mathrm{~T}_{
Rollout 60, Act: -2.694
f_{0}=492 \mathrm{~Hz}$. The sound is reflected by a large
Rollout 97, Act: -2.710
\nGiven the decomposition reaction \(2 \mathrm{~N}_{2} \mathrm{O}_{5
Rollout 54, Act: -2.730
(OH−)\n3. Propionate (CH2CH2COO−)\n4. Methanol
Rollout 97, Act: -2.746
at an initial pressure of $1 \mathrm{~atm}$. After $\mathrm{Y}
Rollout 48, Act: -2.753
the given hex data is: 4500 0034 B612
Rollout 97, Act: -2.793
user\nThe decomposition reaction $2 \mathrm{~N}_{2} \mathrm{O}_{5
Rollout 26, Act: -3.010
2}$ at $298 \mathrm{~K}$ and 1 atm into $\mathrm{
Rollout 27, Act: -3.016
10^{-17} \mathrm{~J}$, calculate the rms speed of the Be
Rollout 48, Act: -3.174
) in the IP header: 4500 0034 B612

Layer 12

GATE_PROJ

Top 16 Positive Activations
Rollout 90, Act: 2.044
9*9=9,081, so congruence holds. Perfect.\n\nThus, for
Rollout 85, Act: 1.989
Since (0,0) is 0, so possible values for the other two are 1 and
Rollout 90, Act: 1.951
1009 is 897, so N≡-M*897 mod1
Rollout 62, Act: 1.916
Yes, 11*173. So two primes.\n\nm=45:20
Rollout 62, Act: 1.908
22 is 11*2. So the result is 473=11
Rollout 62, Act: 1.886
*181=1991. So two primes.\n\nHmm, tedious. So maybe there
Rollout 3, Act: 1.883
which is 300 - 2, so subtract 66.512 *
Rollout 90, Act: 1.881
M=1, N=112. So A is 112 (N=1
Rollout 68, Act: 1.875
at x=-2 is k= -delta, so delta positive approaching zero. So f(-2)=
Rollout 48, Act: 1.868
0x50 is 80, so 2048 + 80 +
Rollout 15, Act: 1.861
.\n\nL1: neighbor L2 (on). So exactly one neighbor on. Therefore, turns on.\n\n
Rollout 90, Act: 1.860
112*1=112. So N=112. That's a three
Rollout 56, Act: 1.852
and A2A0 is ��3, so A3A0 = ��3 /
Rollout 64, Act: 1.847
12=2009+3, so 2^{2012} -1
Rollout 62, Act: 1.839
7*79=1343. So two primes again.\n\nm=38:
Rollout 49, Act: 1.834
. If every subspace is σ-compact, so X is σ-compact. Therefore, X is
Top 16 Negative Activations
Rollout 42, Act: -3.178
determining where the double bond forms. Since the central carbon is attached to two rings (cyclobut
Rollout 54, Act: -3.196
accurate—probably better to compare in terms of base strength. Hydroxide is a much stronger base than
Rollout 42, Act: -3.222
is a -CH2OH group attached to a carbon that is connected to both a cyclobutyl
Rollout 54, Act: -3.224
solvated, so worst nucleophile. RCOO− is weakest base, so least solv
Rollout 54, Act: -3.233
1 reaction, the RDS is the formation of carbocation. If that step is RDS, then
Rollout 54, Act: -3.282
weaker base than hydroxide, but maybe RCOO− is more hindered. Wait, but
Rollout 54, Act: -3.293
solvents (from better to worse): RCOO− (carboxylate) < HO−
Rollout 54, Act: -3.297
than the reactant, which lowers the ΔG‡ and speeds up the reaction. If increasing polarity stabil
Rollout 91, Act: -3.303
Wait, again similar steps. Let's go through carefully calculating angle C:\n\nVector CA=(t -1
Rollout 54, Act: -3.310
Propionate is a carboxylate, RCOO−, so each of them except methanol
Rollout 25, Act: -3.327
require numerical integration, but perhaps approximating using the impulse approximation. Assume during encounter, the time is short
Rollout 42, Act: -3.345
utyl and cyclopropyl), the adjacent carbons are part of these rings. Wait, no
Rollout 54, Act: -3.349
for O nucleophiles in water is typically RCOO− < HO− < RO−. Because
Rollout 42, Act: -3.358
a cyclopropyl group. So the central carbon has three substituents: one hydroxyl
Rollout 54, Act: -3.371
might make it less nucleophilic because the lone pairs are harder to access, but in solution, maybe
Rollout 54, Act: -3.658
ic solvents (from better to worse): RCOO− (carboxylate) < HO

UP_PROJ

Top 16 Positive Activations
Rollout 71, Act: 2.084
the ring of integers, different factorizations might give different solutions. But given the equation e2 +3
Rollout 42, Act: 2.067
). Upon departure of water, a **tertiary carbocation** forms at the central carbon, which
Rollout 42, Act: 2.065
leaving group after protonation (water), perhaps E1 is possible. However, since the substrate is a
Rollout 42, Act: 2.039
hydroxyl group, converting it into a good leaving group (water). Then, the elimination would occur
Rollout 42, Act: 2.011
carbocation is at the central carbon (tertiary maybe?), perhaps adjacent to a three-membered ring
Rollout 49, Act: 2.001
, but is it hereditarily Lindelöf? Probably not. If you take a subset that
Rollout 98, Act: 1.987
quence star. Let's think about what spectral type B5 implies. B5 stars are massive. For
Rollout 49, Act: 1.963
is hereditarily Lindelö, regular (Hausdorff) and it is a Her
Rollout 49, Act: 1.961
uncountable and hereditariy Lindelof, then it has a uncountablesubset with
Rollout 14, Act: 1.961
of integrals and use some sort of dual Hölder applications?\n\nWait, let's see. Let's
Rollout 42, Act: 1.960
ocation. Wait, but if the mechanism is E1, then we first get carbocation formation. However
Rollout 42, Act: 1.929
a good acid, it's more likely an E1 mechanism. But sometimes, with poor leaving groups,
Rollout 42, Act: 1.914
groups: two rings and a hydrogen?), so maybe tertiary alcohols are more prone to E1.
Rollout 49, Act: 1.909
, which would contradict the hereditary Lindelöf property (since a discrete uncountable subspace
Rollout 42, Act: 1.903
xyl group is protonated, forming a good leaving group (water). Upon departure of water, a
Rollout 42, Act: 1.887
is possible. However, since the substrate is a tertiary alcohol (the central carbon is attached to three carbon
Top 16 Negative Activations
Rollout 91, Act: -1.634
| ) = 2t / (2 sqrt( (t -1)^2 + t2 ))
Rollout 65, Act: -1.642
Insight:\n\nWhen you split a.rectangle into two Similar.ones, the raito is squares.\n\n So,
Rollout 56, Act: -1.650
increment:\n\nΔx= (-14)/sqrt(559) * sqrt(559
Rollout 91, Act: -1.655
1 } * 2 ) = t / sqrt{2t2 -2t +1 }\n\nSo
Rollout 95, Act: -1.658
50 = BC, which has length5 sqrt(3), irrational,所以也不rational。\n\nThus
Rollout 95, Act: -1.669
the problem’s original BC length is 5 sqrt(3), perhaps there is an error in prior coordination
Rollout 95, Act: -1.671
total length is PkQk= k sqrt(3)/(2*245 )) = (
Rollout 56, Act: -1.712
Moving along the direction ( (-14)/sqrt(559), (11√3)/
Rollout 53, Act: -1.715
x_C ) /2. x_B is sqrt(73 ) /2. So:\n\nx_B
Rollout 53, Act: -1.719
.5: distance from point P(c/3, d/3) to E(0,0
Rollout 95, Act: -1.727
450 )\n\n Distance between is (9 sqrt(671))/36750.\n\n
Rollout 81, Act: -1.732
lying:\n\n(-51 + h2 )sqrt(533 )= 7*sqrt(
Rollout 53, Act: -1.753
midpoint of BC.\n\nCoordinates of B: ( sqrt(73)/2, 0 )\n\nCoordinates of
Rollout 53, Act: -1.762
C.\n\nFirst, points:\n\nA: (-sqrt(73)/2, 0)\n\nE:
Rollout 53, Act: -1.774
2: distance from point P(c/3, d/3) to D( (a +
Rollout 53, Act: -1.841
3 ) /2. So:\n\nx_B sqrt(73 ) /2 �� something. Add

DOWN_PROJ

Top 16 Positive Activations
Rollout 80, Act: 2.729
10^{-14} \mathrm{~J}$ and reaches a flux density of $1
Rollout 25, Act: 2.687
distance of $3541 \mathrm{~km}$ to Earth's surface with a speed of
Rollout 60, Act: 2.679
f_{0}=492 \mathrm{~Hz}$. The sound is reflected by a large
Rollout 27, Act: 2.627
10^{-17} \mathrm{~J}$, calculate the rms speed of the Be
Rollout 17, Act: 2.611
10^{-26} \mathrm{~J} / \mathrm{T}$. Let the
Rollout 4, Act: 2.607
\pm 0.02 \mathrm{~V}$. What is the percentage error in the
Rollout 97, Act: 2.559
user\nThe decomposition reaction $2 \mathrm{~N}_{2} \mathrm{O}_{5
Rollout 26, Act: 2.405
2}$ at $298 \mathrm{~K}$ and 1 atm into $\mathrm{
Rollout 97, Act: 2.310
\nGiven the decomposition reaction \(2 \mathrm{~N}_{2} \mathrm{O}_{5
Rollout 80, Act: 2.218
of radius $1200 \mathrm{~km}$, density $12.8 \
Rollout 44, Act: 2.214
(\mathrm{p}_{1}, \mathrm{~V}_{1}, \mathrm{~T}_{
Rollout 17, Act: 2.195
10^{-11} \mathrm{~m}$ is the Bohr radius. Evaluate the
Rollout 56, Act: 2.190
0A_2 = \sqrt{3} \), and \( A_1A_2 =
Rollout 97, Act: 2.105
at an initial pressure of $1 \mathrm{~atm}$. After $\mathrm{Y}
Rollout 63, Act: 2.078
sides by \( \frac{4}{3} \):\n\n\[\nf(2, 1)
Rollout 84, Act: 2.073
scenario is \( \frac{2}{9} \) of the original wasted amount. Setting up the
Top 16 Negative Activations
Rollout 14, Act: -1.825
+ K/A.\n\nPlug A=2K/ sqrt(J):\n\nFirst term sqrt(J)* sqrt(A)=
Rollout 64, Act: -1.839
z>=0, the number of integer solutions is floor(C) +1, since z=0,
Rollout 49, Act: -1.850
Every regular Lindelöf space is paracompact and fully normal. But we need the converse
Rollout 57, Act: -1.851
the usual topology, and fails to be paracompact or such, which relates to normality in
Rollout 97, Act: -1.856
is\boxed{2.30}。\n\n**Final Answer**\n\boxed{230}\n\n
Rollout 65, Act: -1.882
)/2=41. But subtracting\n\n**Final Answer**\n\boxed{40}\n\n<|im_start|>
Rollout 99, Act: -1.888
573≈(-0.8235)/1.1573≈-0
Rollout 57, Act: -1.904
regular but not normal. \n \n*A. Csaszar*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so
Rollout 65, Act: -1.909
or equal to1995 is equal to ceiling half the number of total divisors since they come
Rollout 62, Act: -1.920
answer is \boxed{132}。\n\n**Final Answer**\n\boxed{132}\n\n
Rollout 57, Act: -1.926
subsets of the real line that are both F-sigma sets in the Euclidean topology but are not separ
Rollout 17, Act: -1.957
e-11)^3 )\n\nThe 4πe-7 is μ0, which is
Rollout 49, Act: -1.972
a Lindelöf space, hence paracompact. And then in a locally compact parac
Rollout 99, Act: -2.088
75 �� (-0.8235)/1.1575 �� -
Rollout 49, Act: -2.101
space is Lindelöf and thus paracompact, but not necessarily second-countable.\n\nAlternatively
Rollout 49, Act: -2.243
act. And then in a locally compact paracompact Hausdorff space, it's the

Layer 13

GATE_PROJ

Top 16 Positive Activations
Rollout 37, Act: 3.317
products, but maybe inapplicable.\n\nWait, think on the original problem, whether the determinant structure can
Rollout 37, Act: 3.292
grouped under congruence modulo n!.\n\nWait, think of it this way: The set of terms in
Rollout 12, Act: 3.255
[P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the
Rollout 63, Act: 3.250
for the three-variable case.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find a function
Rollout 1, Act: 3.245
mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there
Rollout 6, Act: 3.217
hash of α_n. Not sure.\n\nWait, think about P_angel: the original TM M is
Rollout 51, Act: 3.199
)|\ dx. \]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this integral to solve
Rollout 45, Act: 3.196
of $ m$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem where a
Rollout 11, Act: 3.166
1/c = bc + ac + ab. So p can be written as (a + b + c
Rollout 14, Act: 3.118
dx \right) $$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to prove this inequality
Rollout 28, Act: 3.110
of these hyperbolas.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I've got this problem here
Rollout 83, Act: 3.106
each assigned to a vertex. The sum condition applies cyclically, since it's a nonagon, so
Rollout 66, Act: 3.089
. Find $p$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through this
Rollout 17, Act: 3.076
$ (in SI units).\n<|im_start|>assistant\n<|im_start|>think\nAlright, let me try to tackle this problem
Rollout 69, Act: 3.075
Team Round #10*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find positive integers
Rollout 11, Act: 3.070
) + (ab + bc + ca). So p = a + b + c + ab + bc
Top 16 Negative Activations
Rollout 64, Act: -1.455
as high as 2^{2012 ).\n\nBut in reality, for y=0, when
Rollout 81, Act: -1.461
11, yielding approx72.8 degrees?\n\nThis is a conflict.\n\nWait, so seems there
Rollout 65, Act: -1.469
if a= R =1995√.\n\nBut a= sqr 1995:
Rollout 65, Act: -1.470
Therefore, initially this approach only suggests 1 triple.\n\nBut the problem says "ordered triples (a,b
Rollout 20, Act: -1.470
LCM of 43 and 47)?\n\nWait, but need to check carefully. Wait,
Rollout 97, Act: -1.476
3 s, soY=2.30.\n\nBut problem as it is posed. He might assume
Rollout 52, Act: -1.476
, which is way lower than prior -11.\n\nProceeding.\n\n10. k=10
Rollout 33, Act: -1.482
, so third position is4. Fourth position remaining.\n\nSo permutations starting with2,3,4.\n\n
Rollout 98, Act: -1.487
(let's suppose 7 solar masses, maybe). But such a star would end as a supernova
Rollout 0, Act: -1.493
, the formula is 2^{k -1).\n\nHence, for 20! with
Rollout 99, Act: -1.495
x3 is 2.157 approximately.\n\nHowever, since in the problem they say to perform
Rollout 99, Act: -1.498
while mentioning two iterations, maybe they expect another notation.\n\nTherefore, Given this confusion, but given precise calculation
Rollout 3, Act: -1.513
69. Maybe 168 m/s.\n\nBut checking further on speed, considering the problem might
Rollout 26, Act: -1.534
else perhaps one decimal as 90.4.\n\nWait. 61.32 is precise
Rollout 89, Act: -1.535
that children can be parts? No, obviously not.\n\nAlternative Idea: If this is a certain class,
Rollout 7, Act: -1.602
constraints missed. Alternatively, perhaps N=24?\n\nLet me check manually for N=24.\n\n

UP_PROJ

Top 16 Positive Activations
Rollout 35, Act: 2.885
only two integer solutions, R(n) must be designed so that except for n=19 and
Rollout 87, Act: 2.679
a, z = t a. Thus, each from 0 to 1.\n\nBut returning to substitution
Rollout 65, Act: 2.655
/n to be integer, but n must be at least 2 for k<1. Therefore, this
Rollout 65, Act: 2.605
1995). But c must be at least1995, so if a=1
Rollout 20, Act: 2.542
any positive integer. So, if k is allowed to be any integer, then the requirement is that for
Rollout 72, Act: 2.540
of three roots must be real, hence it's necessary that either they are all real, or the imaginary
Rollout 90, Act: 2.532
1009.\n\nBut since M is a single digit, then (-9N) mod10
Rollout 2, Act: 2.527
1000. As997 terms fromk=3 tok=999.
Rollout 19, Act: 2.518
. However, \( k \) must be at least 6 (since \( n \) is a
Rollout 65, Act: 2.513
.e., exponent at least 1, which given that a2 has exponents even, exponents are
Rollout 0, Act: 2.462
? No, that's not correct. Each must have all primes exponents from 20! For
Rollout 90, Act: 2.437
2018, so A must be at least 2018. However, A must
Rollout 15, Act: 2.413
't bound by mod2. But when you track based on rules (lamp is ON tomorrow only if exactly
Rollout 90, Act: 2.384
M, but the problem specifies that M is a single digit, 0-9. Therefore, only
Rollout 20, Act: 2.379
for all positive integers a". When a is allowed to be any positive integer, we have to consider,
Rollout 65, Act: 2.370
Similarly, trying other permutations, but perhaps with each other.\n\nBut thinking again, in terms of cutting along
Top 16 Negative Activations
Rollout 12, Act: -1.923
is monotonically decreasing in the interval $ 0<\nu<1$ . [P
Rollout 55, Act: -1.929
{A}^{2018}.$\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I
Rollout 86, Act: -1.930
2}}f(x)\ dx.$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so the problem is asking
Rollout 18, Act: -1.943
such that $H=H_{xy}$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's
Rollout 69, Act: -2.005
in degrees, compute $a+b$ .\n\n\n*2021 CCA Math Bonanza
Rollout 77, Act: -2.015
Calculate $f(14,52)$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 90, Act: -2.029
are divisible by $2018$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I
Rollout 71, Act: -2.043
a| + |b| + |c|$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 12, Act: -2.043
$ 0<\nu<1$ . [P. Turan]\n<|im_start|>assistant\n<|im_start|>
Rollout 72, Act: -2.092
whose sum is $ \n1. $ \n<|im_start|>assistant\n<|im_start|>think\nAlright, let's
Rollout 71, Act: -2.144
, $b$ , and $c$ . Find $|a| + |b| +
Rollout 72, Act: -2.145
.\n<|im_start|>user\nFind all numbers $ n $ for which there exist three (not necessarily distinct
Rollout 72, Act: -2.147
$ n $ whose sum is $ \n1. $ \n<|im_start|>assistant\n<|im_start|>think\n
Rollout 51, Act: -2.147
sin (x-\theta)|\ dx. \]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 1, Act: -2.170
vectors in $ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 86, Act: -2.261
{\pi}{2}}f(x)\ dx.$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so the

DOWN_PROJ

Top 16 Positive Activations
Rollout 30, Act: 2.087
1 ).\n\nTherefore, (b_n +1 ) term can be written as a ratio of a_{n
Rollout 3, Act: 1.985
Let me make sure.\n\nAccording to Maxwell-Boltzmann distribution, the mean speed is the average of
Rollout 73, Act: 1.954
- 2046=16388.\n\nTherefore, going back to N:\n\nN
Rollout 87, Act: 1.928
numerical constants, so answer would be expression like \( \frac{\rho a^2}{\epsilon_
Rollout 64, Act: 1.893
from numerator:\n\n=2^{2009}[8 +a1*(1 -2/2
Rollout 3, Act: 1.810
(average) speed derived from the Maxwell-Boltzmann distribution:\n\n\[\nv = \sqrt{\
Rollout 26, Act: 1.684
s)$:\n \[\n \Delta H_f^\circ (\mathrm{HgO}) =
Rollout 42, Act: 1.646
ocation formation. However, if the carbocation is adjacent to strained rings, like cyclopropane,
Rollout 91, Act: 1.641
= (t +1)/sqrt(2 t2 + 2 t +1 )\n\nSimilarly, sin
Rollout 10, Act: 1.634
theory here.\n\nRecall by Legendre's three-square theorem, a number n is expressible as a
Rollout 3, Act: 1.624
8 K (unless stated otherwise), perhaps answer around \boxed{169} m/s. But
Rollout 62, Act: 1.621
121: compute expression=1212-121 +11=14
Rollout 75, Act: 1.618
depends on p.\n\nBut by Legendre's three-square theorem, a positive integer n can be expressed as
Rollout 99, Act: 1.608
0.8027467 z2 -0.7993134
Rollout 81, Act: 1.590
angle between AB and AC, so the actual angle's tangent is positive, so perhaps the absolute value applies
Rollout 14, Act: 1.590
HS=1 and RHS=4. Something's not adding up here. Wait, but the question states
Top 16 Negative Activations
Rollout 95, Act: -2.871
)/15 + 0 = (9 sqrt(671))/36750.\n\n
Rollout 95, Act: -2.871
( -2/3)^2 + ( sqrt(671)/3 )^2 )=
Rollout 53, Act: -2.892
)/2))= -27/(2 sqrt(73))= multiplied by sqrt(73
Rollout 95, Act: -2.914
) to C(25/3, sqrt(671)/3).\n\n vector components:2
Rollout 53, Act: -2.918
73 ) = -27/(2 sqrt(73)) ) Rationallze: this
Rollout 72, Act: -2.920
^2)=1 + 2*(-(sqrt(5)+1)/4)=1 - (sqrt
Rollout 53, Act: -2.928
as:\n\nc= -27/(2 sqrt(73)), and in equation3:\n\nc2
Rollout 95, Act: -2.934
450 )\n\n Distance between is (9 sqrt(671))/36750.\n\n
Rollout 95, Act: -2.955
) to C(25/3, sqrt(671)/3),\n\n So AQk is
Rollout 95, Act: -2.958
3675 )= (15 sqrt(3))/(2*3675 )=
Rollout 95, Act: -2.959
7/3= (-2/3, sqrt(671)/3 ) ,length sqrt(
Rollout 95, Act: -2.964
up to (25/3 a, sqrt(671)/3 a ), adnd length
Rollout 95, Act: -3.017
the problem’s original BC length is 5 sqrt(3), perhaps there is an error in prior coordination
Rollout 53, Act: -3.025
, then c=-27/(4*(sqrt(73)/2))= -27/(
Rollout 95, Act: -3.043
)/3675= (15 sqrt(3))/(2*3675)= sqrt
Rollout 95, Act: -3.081
这样的话,k=490*(c/sqrt(3))*sqrt(3 )=490

Layer 14

GATE_PROJ

Top 16 Positive Activations
Rollout 60, Act: 1.527
98\n\nTherefore, 492 * (332/328) =
Rollout 25, Act: 1.385
_final = 0\n\nSo,\n\n2 μ_final r_p_final = v2 (r2 - r
Rollout 29, Act: 1.339
. Since it's open, gauge pressure is rho*g*h. Meanwhile, in the handle, the pressure
Rollout 29, Act: 1.334
pressure at the bottom of the handle would be rho*g*H. Then the gauge pressure would be rho
Rollout 29, Act: 1.316
*H. Then the gauge pressure would be rho*g*H - atmospheric pressure. But since the main
Rollout 60, Act: 1.287
, f' = 492 Hz * (330 + 2)/330
Rollout 78, Act: 1.280
_w + 60\n\nSo:\n\nL / (v_w - 20) = 1
Rollout 78, Act: 1.272
40) in the equation:\n\nL / (v_w - 20) = 1
Rollout 60, Act: 1.269
v - v_car) ) = f0 * (v + v_car)/(v - v_car)\n\n
Rollout 29, Act: 1.261
pressure at the bottom of the handle would be rho*g*H, but if the main jug is connected
Rollout 29, Act: 1.255
main jug's bottom has a gauge pressure of rho*g*h, while the handle's bottom (if considered
Rollout 25, Act: 1.254
\n\nThus,\n\nΔd≈ (1/2)a_mean t2 = (1/2)*( G
Rollout 29, Act: 1.253
pressure of the sealed air in the handle plus rho*g*H. Since the pressure at the interface between
Rollout 29, Act: 1.235
the bottom of the main jug would still be rho*g*h (gauge pressure, since it's open
Rollout 13, Act: 1.228
1936\n\nTherefore, A_annual = 6000 * 1.5
Rollout 29, Act: 1.224
situation where the gauge pressure at the bottom is rho*g*h regardless of the higher handle level? If the
Top 16 Negative Activations
Rollout 52, Act: -3.836
028 is the maximum k. So as such, if we think this through, n is (
Rollout 68, Act: -3.849
be negative (since k negative, x negative: thus kx positive). But our domain was indeed x
Rollout 89, Act: -3.851
be split. Which would have been an issue. Thus, the answer. Maybe the problem does not specify
Rollout 15, Act: -3.852
n=1=2^0. Let's think of our n= 2^k.\n\nFor
Rollout 89, Act: -3.854
.\n\nBut the conventional question would likely state integers. Hence, with that given, if you take T
Rollout 20, Act: -3.855
but need to check carefully. Wait, let's think:\n\nBut for modulus 43: n �
Rollout 75, Act: -3.860
16 +16 +0 again.\n\nThis leads me to think that there is an edge case when
Rollout 62, Act: -3.860
values of m where the quadratic expression is larger, hence more likely to be composite with multiple factors.\n\nLet
Rollout 72, Act: -3.868
"number n" could refer to positive integers. Thus, should we answer as in the problem's context
Rollout 85, Act: -3.940
. For example, let's suppose we start alternately labeling.\n\nFrom (0,0)=0,
Rollout 89, Act: -3.947
it's impossible for the primary school math problem. Thus, perhaps made a mistake. But in lieu of
Rollout 41, Act: -3.950
treating as four-digit with leading zeros: can also think of different way.\n\nTotal counted when allowing leading zeros
Rollout 37, Act: -3.953
-a1) and for each variable, but hmm.\n\nWait, perhaps there is more structure here.
Rollout 93, Act: -4.068
considered as some critical points. Hmm. I'm thinking of using Orthogonal Polynomials machinery here but I
Rollout 97, Act: -4.084
via ODE and stoich integration. Hmm—thus integrating via this approach to total pressure gets same result
Rollout 68, Act: -4.119
) ] / 2.\n\nBut let's think alternatively. For when quadratic has two roots, maybe we

UP_PROJ

Top 16 Positive Activations
Rollout 99, Act: 2.239
was approximately1.1573, adding to x2=1, we getx3≈
Rollout 64, Act: 2.171
} +1 )\n\nBut likely. Or in other terms, could write this as9*4*2
Rollout 74, Act: 2.140
. So yes, that seems right. So combining with the fore-and-aft components, which when resolved
Rollout 97, Act: 2.135
1.45 atm. But if the above is the case (which a first-order decay of N
Rollout 89, Act: 2.077
the exact decimal 0.4, but that also would require T such that 0.4*T
Rollout 65, Act: 2.062
k*c, c = k*a, but that also seems tricky.\n\nAlternatively, maybe it's possible that
Rollout 92, Act: 2.054
side as the object). \n\nSo, plugging into the lens formula: 1/f = 1
Rollout 1, Act: 2.040
problem is over the Hilbert space, so thinking as in R^n, for which we know that even
Rollout 57, Act: 2.040
C, then since the complement of C is open and contains x, there exists a measurable set A (
Rollout 81, Act: 2.030
(angle BAD)*tan(angle CAD)].\n\nThat's from the tangent addition formula.\n\nGiven angle BAC=
Rollout 39, Act: 2.019
89 (which we calculated earlier). So inserting in a number into the end, yes, if you
Rollout 67, Act: 2.006
. So if we proceed...\n\nBut this is consistent with our calculations. So yes, then the eighth term
Rollout 11, Act: 2.005
bc+ ca. compute p2. Maybe that expands to something that can be combined with q, but
Rollout 1, Act: 1.965
equations intersect at a common y.\n\nBut the problem now is to show that the system of equations ||x
Rollout 46, Act: 1.962
in the rotating frame. Alternatively, maybe it's from a Lagrangian where the variables are the rotating
Rollout 91, Act: 1.957
) = ( t -1 , t )\n\nSo now, we can compute the angles at B and C
Top 16 Negative Activations
Rollout 83, Act: -1.957
Hence, the labels must form cycles under rotation of k positions.\n\nTherefore, the cycle length for a rotation
Rollout 15, Act: -1.959
k, then any initial state will reach zero after k steps.\n\nTherefore, if we can show that M
Rollout 83, Act: -1.976
cycles of length 9/d. So rotations by k where gcd(9,k)=d will have cycle
Rollout 83, Act: -1.978
But each group must have three elements assigned the same residue. Remember, residue counts: 0,1
Rollout 93, Act: -1.986
1007 for j=1007.\n\nSimilarly, j=2014:
Rollout 16, Act: -1.988
Each sapling independently succeeds with probability 1/2 each day. So the process continues until all six
Rollout 93, Act: -1.989
choose1007 times 1007 over2^{2014}}, plus2
Rollout 83, Act: -1.990
Wait no: residue groups are assignments of digits into residue classes, which are fixed. So even if a
Rollout 56, Act: -1.992
K such that her pasture never exceeds an area of K".\n\nGrammatically: after each extension, "
Rollout 16, Act: -1.994
Then the probability that it hasn't been converted after t days is (1/2)^t. Therefore
Rollout 14, Act: -2.003
the maximum possible ratio is approximately 2.25/8≈0.28. Wait,
Rollout 83, Act: -2.034
residues needed:\n\nGroup0 has three positions, assigned residue a --> so that's three instances of a residue
Rollout 93, Act: -2.034
a root of unity of order 1007/(k,1007). Wait,
Rollout 16, Act: -2.045
(sapling) independently with probability 1/2.\n\nBut is there an analogous formula here? Let
Rollout 65, Act: -2.118
Original definition: since c is derived from a via c=19952/ a. If
Rollout 93, Act: -2.154
007 implies j ��1007 mod 1007, i.e.,

DOWN_PROJ

Top 16 Positive Activations
Rollout 98, Act: 2.577
, RGB star (1.5Msun)** \n - **Invalid** \n - **Population
Rollout 27, Act: 2.453
/2))^(2/3)\n\nwhere ��(3/2) is Apéry's
Rollout 27, Act: 2.361
� is the reduced Planck constant, and ��(3/2) is the Riemann
Rollout 3, Act: 2.270
square speed, which is related to temperature and molar mass. Wait, but the question specifically says "
Rollout 27, Act: 2.254
m k_B) ) * (1 / ��(3/2))^(2/3)\n\n
Rollout 27, Act: 2.230
So m = 9 * 1.6605e-27 �� 1
Rollout 3, Act: 2.195
the temperature in Kelvin, and M is the molar mass. Then there's the average speed, which
Rollout 27, Act: 2.173
units). 1 u = 1.6605e-27 kg. So m
Rollout 27, Act: 2.131
\text{u} = 1.6605 \times 10^{-27
Rollout 17, Act: 2.129
, where $a_{0}=5.29 \times 10^{-11} \
Rollout 22, Act: 2.127
R is the gas constant and M is the molar mass. Wait, really? So there's another
Rollout 22, Act: 2.126
P)\n- \( M \) is the molar mass of hydrogen gas (\( 0.0
Rollout 3, Act: 2.078
Kelvin,\n- \( M \) is the molar mass of radon in kg/mol.\n\n**Steps
Rollout 22, Act: 2.070
in grams, confused with kg.\n\nWait the molar mass is 2.016 g/mol
Rollout 93, Act: 2.066
connected to sum= 2^{2n -1} (1 + C(2n,n)
Rollout 22, Act: 2.043
, each molecule has two hydrogen atoms. The molar mass of hydrogen is approximately 2.01
Top 16 Negative Activations
Rollout 95, Act: -1.894
3*(k/2450), sqrt(671)/3*(k/24
Rollout 95, Act: -1.894
up to (25/3 a, sqrt(671)/3 a ), adnd length
Rollout 95, Act: -1.899
3)k/2450, sqrt(671)/3 *k/24
Rollout 93, Act: -1.900
2... So even the first term wouldn't decay that much? Wait, but if those approximations apply
Rollout 41, Act: -1.902
. From this, subtract the 9 numbers with repeating digits. So 90 - 9 =
Rollout 95, Act: -1.910
450)*(-2/3, sqrt(671)/3), which is in the
Rollout 42, Act: -1.910
central structure is:\n\nCyclobutyl-C(OH)H-Cyclopropyl\n\nBut
Rollout 38, Act: -1.920
[31]2 (x2 + W2)\n\nCompute left side:\n\n625(W +
Rollout 95, Act: -1.927
) to C(25/3, sqrt(671)/3),\n\n So AQk is
Rollout 38, Act: -1.933
= 6.2 sqrt(x2 + W2)\n\nLet me express 6.2 as
Rollout 9, Act: -1.938
and GM.\n\nEquation for GE (from G(10 - h,5) to E(
Rollout 38, Act: -1.959
2 = 961(x2 + W2)\n\nDivide both sides by (W + x
Rollout 1, Act: -1.971
d / sqrt(2) + d / sqrt(2) = d sqrt(2} ~1
Rollout 95, Act: -1.973
90*(c/sqrt(3))*sqrt(3 )=490*c,其中c
Rollout 71, Act: -1.994
d, and then 3d2 + e2=2011. Then k=2
Rollout 3, Act: -2.151
the mean speed is called average speed and given by that formula with 8/π, while the rms

Layer 15

GATE_PROJ

Top 16 Positive Activations
Rollout 45, Act: 1.439
that $ f(1) \equal{} 1, f(2) \equal{}
Rollout 49, Act: 1.404
subset property? I think so. Let me check this.\n\nIf X is hereditarily Lindelö
Rollout 1, Act: 1.368
are mutually orthogonal after scaling.\n\nWait, let me recall that if you have a regular simplex in R^n
Rollout 9, Act: 1.358
}\n\nWait, that seems correct.\n\nLet me check this.\n\nGiven that the area is 100
Rollout 88, Act: 1.354
referring to weights. Hmm.\n\nWait, let me check again: "What is the total number of weights
Rollout 33, Act: 1.328
(6 - km)!.\n\nWait, let me check with a concrete example. Take A_1 �
Rollout 5, Act: 1.326
the two bases. Wait, no, let me check. \n\nI remember that for areas divided by a
Rollout 39, Act: 1.263
total? Wait, no. Wait, let me check. If the first term is 1 digit,
Rollout 18, Act: 1.258
need x to be positive.\n\nFirst, let me recall that in Z2, any subgroup is isomorphic
Rollout 15, Act: 1.257
mathematics assistant.\n<|im_start|>user\nLet $n\geq1$ be a positive integer.
Rollout 90, Act: 1.249
itself is 4 digits). Wait, let me think again.\n\nWait, hold on. 20
Rollout 45, Act: 1.242
q mx^2\plus{}(5m\plus{}1)x\plus{}4m$
Rollout 1, Act: 1.234
two distinct vectors are orthogonal.\n\nFirst, let me recall that an orthonormal system requires two things:
Rollout 74, Act: 1.233
to the magnetic fields causing the deviation. Let me check my notes. Oh, right! In the context
Rollout 65, Act: 1.214
c) satisfy this condition.\n\nFirst, let me recall what similar prisms mean. If one prism is
Rollout 66, Act: 1.213
to find the perimeter p.\n\nAlright, let me recall that in a right-angled triangle, the altitude
Top 16 Negative Activations
Rollout 42, Act: -3.537
ring. But cyclopropane is three-membered; forming a double bond adjacent to it would create
Rollout 2, Act: -3.548
00\n\nSimilarly,10^k -1= -1 mod1000 fork>=
Rollout 42, Act: -3.559
propane) and the other is 4-membered (cyclobutane), linked through a
Rollout 70, Act: -3.564
^4 -1=2400-1=2400) be a 7
Rollout 69, Act: -3.586
ometric identity. 2 cos 2 theta -1 is equal to cos 2 theta? Wait,
Rollout 42, Act: -3.591
a double bond such that one ring remains three-membered and the other part is a four-membered.\n\n
Rollout 64, Act: -3.611
)\n\nSince 2^{2009}-1 is large, the fraction is just under 8
Rollout 15, Act: -3.612
Because n=1 is 2^1 -1 =1, and then maybe 3,
Rollout 64, Act: -3.613
But 2^{2012} -1 -K y is already an integer because both M
Rollout 64, Act: -3.627
while 2^{2009} -1 is positive for a2.\n\nSo we can think
Rollout 42, Act: -3.629
ring. Cyclopropane is 3-membered; opening could create a 4-, 5
Rollout 42, Act: -3.629
rings and a central carbon, getting a six-membered ring seems unclear. Option C is hexahy
Rollout 15, Act: -3.632
=3, which is 2^2 -1: no, since n=3 was bad.
Rollout 42, Act: -3.637
spiro union where one ring is 3-membered (cyclopropane) and the other
Rollout 2, Act: -3.648
;\n\nk=2,10^2 -1=99\n\nk=3, so1
Rollout 64, Act: -3.659
Since 2^{2009} -1 = a very large number, 7/(2

UP_PROJ

Top 16 Positive Activations
Rollout 77, Act: 2.110
Calculate $f(14,52)$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 90, Act: 2.109
are divisible by $2018$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I
Rollout 11, Act: 2.096
values of the ordered pair $(p,q)$ .\n\n*David Altizio*\n<|im_start|>assistant\n<|im_start|>think
Rollout 39, Act: 2.093
Find the sum of the digits of $N$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, let me
Rollout 78, Act: 2.084
relatively prime positive integers. Find $m+n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me
Rollout 84, Act: 2.082
$ as much beverage. Find $m+n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 9, Act: 2.054
to $EM$ in $\triangle GEM$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 59, Act: 2.049
prime positive integers. Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so the
Rollout 70, Act: 2.041
-10 number that is twice $N$ . For example, $51$ is a
Rollout 5, Act: 2.008
exceed $x^2/100$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's
Rollout 16, Act: 2.007
. Find $100m+n$ .\n\n*Proposed by Yannick Yao*\n<|im_start|>
Rollout 8, Act: 1.991
prime positive integers. Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's
Rollout 50, Act: 1.986
relatively prime positive integers. Find $m+n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me
Rollout 20, Act: 1.946
factors in the prime factorization of $n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 15, Act: 1.934
a helpful mathematics assistant.\n<|im_start|>user\nLet $n\geq1$ be a positive
Rollout 45, Act: 1.934
a helpful mathematics assistant.\n<|im_start|>user\nLet $ f: Z \to Z$ be such
Top 16 Negative Activations
Rollout 44, Act: -1.913
temperature changes. Because in the example where I took reversible isothermal, but if instead the process is reversible
Rollout 11, Act: -1.917
1. Then, p = x + y + z + xy + yz + zx. And q
Rollout 11, Act: -1.925
r_a e^{iθ_a}, b = r_b e^{iθ_b}, c = r
Rollout 11, Act: -1.926
for some integer k. Hmm, as modulus: r_a r_b r_c = 1.\n\nBut
Rollout 11, Act: -1.933
�, b = a (so same modulus and argument), c =1/a2. Then compute p
Rollout 11, Act: -1.935
consider all three variables with the same modulus, say r, but different arguments. Then since product is
Rollout 69, Act: -1.948
at the first term:\n\nIn term_0, angle = 10°, so sin2(
Rollout 69, Act: -1.950
(10°)\n\nIn term_2, angle=10°/9: sin2(
Rollout 11, Act: -1.955
+1/r3.\n\nNote that if R = r3 +1/r3, then cosθ =
Rollout 69, Act: -1.968
i−1})\n\nLet me denote that phi = theta_{i−1}\n\nTherefore, term_i =
Rollout 11, Act: -1.995
such that a = \overline{b}, c is real? But then c must not be real
Rollout 37, Act: -2.056
,2,...,2015, and k_j= j for simplicity. Then matrix A is
Rollout 91, Act: -2.083
D. Since BD is equal to DC, but angles opposite to them might not be equal. Wait,
Rollout 69, Act: -2.095
=0 to infinity, which corresponds to terms with angles 10°, 10/3
Rollout 11, Act: -2.117
1. Since 1 has modulus 1 and argument 0, so modulus r_a r_b r
Rollout 69, Act: -2.138
(30°)\n\nIn term_1, angle =10°/3: sin2(

DOWN_PROJ

Top 16 Positive Activations
Rollout 64, Act: 2.043
)*2^n = k*2^{n-1}, which is integer for n ≥1. Therefore
Rollout 17, Act: 2.036
θ.\n\nLet me compute the polar integral. Let u = cosθ, du = -sinθ d
Rollout 87, Act: 2.031
to standard integrals:\n\nThe integral �� du / sqrt(u2 + c) = sinh−1
Rollout 87, Act: 2.018
:\n\nThe integral ��0^a dx / sqrt(x2 + y2 + z2) =
Rollout 49, Act: 2.013
spaces that are not regular, called Tychonoff corkscrew or similar examples. But if the
Rollout 87, Act: 2.006
integral, I can use substitution. Let’s set u = x, so the integrand is 1
Rollout 73, Act: 2.004
- (n+1)2^{n+1} + n2^{n+2} =
Rollout 87, Act: 1.999
standard integrals:\n\nThe integral �� du / sqrt(u2 + c) = sinh−1(u
Rollout 87, Act: 1.963
� r ln r dr. Let’s do substitution u = ln r, dv = r dr. Then
Rollout 51, Act: 1.947
\ dx. \]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this integral to solve:
Rollout 44, Act: 1.943
20*(20^(1/3))^2≈20*(2.714)^
Rollout 87, Act: 1.922
.\n\nPerhaps instead do a substitution. Let’s set u = y, v = z. Not helpful.
Rollout 86, Act: 1.904
with x such as x*sinx*cosx^2. That was case9≈0.22
Rollout 87, Act: 1.881
integrals:\n\nThe integral �� du / sqrt(u2 + c) = sinh−1(u /
Rollout 12, Act: 1.875
_a^b \frac{dt}{\sqrt{(t - a)(b - t)(t
Rollout 57, Act: 1.857
. Csaszar*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that if we
Top 16 Negative Activations
Rollout 54, Act: -2.586
conjugate base, ethanethiolate (HS− or CH2CH2S−), is
Rollout 54, Act: -2.637
to worse): RCOO− (carboxylate) < HO− < RO−. Wait
Rollout 54, Act: -2.663
stronger acids than alcohols, so thiolate is a weaker base? Wait, pKa of
Rollout 54, Act: -2.671
S has a pKa around 7. So HS− is a weaker base than OH−, right
Rollout 20, Act: -2.672
)^× is cyclic, there exist primitive roots where ord_m(p) = m -1, and elements
Rollout 54, Act: -2.737
Ka ~10 of ethanethiol → HS-R pKa around 10). Wait,
Rollout 54, Act: -2.746
iolate. Propionate is a carboxylate, RCOO−, so each of them
Rollout 21, Act: -2.775
, the functional equation given is similar to Cauchy but incremental. Let’s see: f(x +
Rollout 54, Act: -2.803
. Wait, no. Typically, carboxylates are less nucleophilic than hydroxide,
Rollout 54, Act: -2.845
solution:\n\nIn SN2 reactions, the thiolate ion (e.g., SH−) is a
Rollout 54, Act: -2.937
of H2S is around 7, so HS− is a stronger base than OH−? Wait
Rollout 54, Act: -2.953
more polarizable. For example, in water, HS− is a better nucleophile than OH− despite
Rollout 54, Act: -3.041
propionate (which it is—carboxylate resonance), then propionate would have a much lower
Rollout 54, Act: -3.093
That makes sense.\n\nEthanethiolate (HS− or RS−): Sulfur is larger
Rollout 54, Act: -3.127
worse): RCOO− (carboxylate) < HO− < RO−. Wait,
Rollout 54, Act: -3.470
3OH)\n5. Ethanethiolate (HS− or maybe C2H5S−?

Layer 16

GATE_PROJ

Top 16 Positive Activations
Rollout 11, Act: 4.026
the same modulus.\n\nAlternatively, two variables with reciprocal modulus and one with modulus not 1.\n\nBut overall
Rollout 11, Act: 3.815
= \overline{1/b}, leading to modulus |a| =1/|b | ;
Rollout 11, Act: 3.686
real. But since a is neither real nor of modulus 1, this is not the case.\n\nWait
Rollout 11, Act: 3.657
|=1/k2 <1. So none have modulus 1, which is okay. Then, set
Rollout 11, Act: 3.632
= ��, b = a (so same modulus and argument), c =1/a2. Then
Rollout 11, Act: 3.610
1. Let’s consider them with moduli reciprocal-related. Like |a|=k>1,
Rollout 11, Act: 3.574
?\n\nAlternatively, recall the problem states that none have modulus 1. If we let’s take a,
Rollout 11, Act: 3.572
. Hence, all three variables cannot have the same modulus.\n\nAlternatively, two variables with reciprocal modulus and one
Rollout 11, Act: 3.570
1. But problem states that none of them have modulus 1. Hence, |a| and |
Rollout 11, Act: 3.539
modulus 1. Let’s consider them with moduli reciprocal-related. Like |a|=k>1
Rollout 72, Act: 3.521
So, maybe the sum can be arranged to be real. Indeed, for example, 1 + �
Rollout 11, Act: 3.495
product is 1. But a and b have modulus r ≠1, c has modulus 1/(
Rollout 52, Act: 3.484
positive 2014. Let's check partial sum.\n\nBut in reality, the sum from -
Rollout 11, Act: 3.473
}, c= e^{-i2θ}. But modulus of a and b is 1, which is
Rollout 11, Act: 3.451
none of a, b, c are real or modulus 1. Let’s consider them with moduli
Rollout 11, Act: 3.428
their product is 1, so if two have modulus greater than 1 and one less, but their
Top 16 Negative Activations
Rollout 91, Act: -1.802
2} = sqrt{2t2 - 2t +1 }.\n\nSo sin(theta_C )
Rollout 36, Act: -1.812
- 2sx - 2sy - 2sz + 3s2 = 90
Rollout 36, Act: -1.813
0 - s2 - 2sy - 2sz + 3s2 = 90
Rollout 66, Act: -1.815
6ab - 6b2 - 6b c = 0\n\nHmm, perhaps group
Rollout 97, Act: -1.822
1}\right) = 2 \times 5 \times 10^{-4} \times
Rollout 28, Act: -1.825
2 / (2h -a) - 2 h b2 / (2k -b)\n\n
Rollout 28, Act: -1.849
\frac{2A a^2 + 2C b^2}{ab} = 2
Rollout 36, Act: -1.862
- 2sx - 2sy - 2sz + 3s2 = 90
Rollout 36, Act: -1.865
- 2sx - 2sy - 2sz + 3s2 = 90
Rollout 36, Act: -1.872
- 2sx - 2sy - 2sz + 3s2 = 90
Rollout 28, Act: -1.874
hk - 2a b2 h - 2a2 b k + a2 b2 +
Rollout 28, Act: -1.881
hk - 2a b2 h - 2a2 b k + a2 b2 =
Rollout 28, Act: -1.946
:\n\n\( (-2A a^2 - 2C b^2) + B ab =
Rollout 36, Act: -1.954
- s2) - 2sx - 2sy - 2sz + 3s2
Rollout 36, Act: -2.019
sx - s2 - 2sx - 2sy - 2sz + 3s2
Rollout 28, Act: -2.096
a2 (2k -b) - 2 h b2 (2h -a )\n\nExpand

UP_PROJ

Top 16 Positive Activations
Rollout 16, Act: 2.163
each remaining sapling has a 1/2 chance to convert. Let me model this as a Mark
Rollout 36, Act: 2.141
:\n\nBut I need to be careful. So the compute showed that S=36,864
Rollout 16, Act: 2.115
and each sapling has a 1/2 chance of turning into a perfect tree that day. Once
Rollout 89, Act: 2.096
.\n\nBut unless specified. But considering that school problems typically avoid this. So possible dupe:\n\nGirls
Rollout 16, Act: 2.068
all remaining saplings have a 1/2 chance of converting. So each day, multiple can be
Rollout 16, Act: 2.064
still a sapling has a 1/2 chance to convert. The process is similar to independent trials
Rollout 22, Act: 2.042
"compute the mean molecular speed...", given that's under unspecified temperature, both are reasonable. But in physics
Rollout 97, Act: 2.025
≈2.3。but given real exams problems often demand which, I'm torn.\n\nThe confusion arises
Rollout 39, Act: 2.019
perhaps another example.\n\nBut maybe stepping back. The principal part of the logic is that, given that each
Rollout 29, Act: 1.998
explicitly asks for the gauge pressure, which is formulaically P = rho g h for an open container,
Rollout 54, Act: 1.962
ethiolate. Propionate is a carboxylate, RCOO−, so each of
Rollout 37, Act: 1.961
when considering as variables, the symmetric polynomial S is such that n! divides it. Wait, but how
Rollout 64, Act: 1.957
(2^{3} -1 )\n\nWait, long division. Since 2012=2
Rollout 25, Act: 1.952
Since gravitational force depends on both masses (Newton's law of gravitation), but in the case of the
Rollout 27, Act: 1.951
Beryllium-9 has 4 protons and 5 neutrons, so total spin integer
Rollout 60, Act: 1.942
.\n<|im_start|>user\nA stationary source emits sound of frequency $f_{0}=492 \mathrm
Top 16 Negative Activations
Rollout 86, Act: -1.812
(x2)(1/2) sin2x | from0 to pi/2 - ��
Rollout 86, Act: -1.827
returning, the integral ��x2 cos2x dx= -∫x sin2x dx=
Rollout 86, Act: -1.838
But when we compute ��x2 cos2x dx= - ��x sin2x dx
Rollout 86, Act: -1.841
/4. So ��x2 cos2x dx= -pi/4 ?\n\nBut that can
Rollout 69, Act: -1.849
theta)\n\n= cos(theta) - 2 cos(theta) sin^2 theta - 2 sin^
Rollout 86, Act: -1.855
+ (1/2) ��cos2x dx ]\n\nCalculating this:\n\n- [ - (
Rollout 86, Act: -1.858
+ (1/2) �� cos2x dx ]\n\n= - [ - (1/2
Rollout 86, Act: -1.862
+ C.\n\nThus, ��sin^3x dx= -cosx + (1/3
Rollout 86, Act: -1.874
x dx, then ��x2 cos2x dx = (1/2)x^2 sin
Rollout 51, Act: -1.879
= sin8x cos8θ + cos8x sin8θ\n\nTherefore, the integral becomes:\n\n
Rollout 86, Act: -1.881
Therefore, the original ��x2 cos2x dx = pi/4.\n\nTherefore, returning to
Rollout 86, Act: -1.883
.\n\nSecond integral: ��x2 cos2x dx. Integration by parts.\n\nLet’s let u
Rollout 86, Act: -1.899
+ (1/2) �� cos2x dx )?\n\nWait no, that earlier process gave
Rollout 86, Act: -1.920
4)\n\nTherefore, ��x2 cos2x dx = -∫x sin2x dx =
Rollout 86, Act: -1.929
+ (1/2) �� cos2x dx.\n\nEvaluate terms:\n\nAt pi/2:
Rollout 0, Act: -1.933
are related to the divisors of 20! where each pair consists of two coprime numbers

DOWN_PROJ

Top 16 Positive Activations
Rollout 54, Act: 2.797
base than hydroxide, but maybe RCOO− is more hindered. Wait, but regardless
Rollout 54, Act: 2.629
(alkoxide) < HO− < RCOO−. Because RO− is strongest base, most
Rollout 54, Act: 2.623
ionate is a carboxylate, RCOO−, so each of them except methanol has
Rollout 54, Act: 2.603
solvents (from better to worse): RCOO− (carboxylate) < HO−
Rollout 54, Act: 2.603
), propionate (3):\n\nPropionate (RCOO−) is the weakest base, so in polar
Rollout 54, Act: 2.586
being strong base)\n\n3. Propionate (RCOO−, resonance stabilized, weakest base among the charges
Rollout 54, Act: 2.560
polar protic. So if propionate (RCOO−) is weaker base than hydroxide,
Rollout 54, Act: 2.522
vated, so worst nucleophile. RCOO− is weakest base, so least solvated
Rollout 54, Act: 2.506
O nucleophiles in water is typically RCOO− < HO− < RO−. Because even
Rollout 12, Act: 2.263
int_1^{\frac{1}{\nu}} \frac{dx}{\sqrt{(x
Rollout 46, Act: 2.108
angular velocity ��. The acceleration in the inertial frame is a_inertial = a_rotating
Rollout 46, Act: 1.995
frame) is related to the acceleration in the inertial frame a by:\n\na' = a -
Rollout 54, Act: 1.935
vents (from better to worse): RCOO− (carboxylate) < HO− <
Rollout 46, Act: 1.926
kinetic energy is different. The velocity in the inertial frame is v_inertial = v_rotating
Rollout 54, Act: 1.894
hydroxide is a better nucleophile than acetate (which is a weaker base). Because perhaps despite
Rollout 12, Act: 1.886
we had x from 1 to 1/ν, substitution t=νx gives t from ν
Top 16 Negative Activations
Rollout 23, Act: -2.264
considers the alpha decay half life and neglects the beta decay, they calculate answer (A). But the
Rollout 23, Act: -2.269
.\n<|im_start|>user\nImagine a radioactive nuclei X(Z,A) can decay into Y(Z-2, A
Rollout 23, Act: -2.281
0385 s−1\n\nSimilarly, beta decay: T_beta=0.098
Rollout 23, Act: -2.288
both decays. Someone might have ignore the competing beta decay. Let's check that.\n\nIf there was
Rollout 23, Act: -2.301
need to consider the decay constants for both alpha and beta decay, sum them up for the total decay constant
Rollout 62, Act: -2.305
Euler's famous polynomial n^2 +n +41, which generates primes for n=0 to
Rollout 27, Act: -2.380
mmm.\n\nWait, perhaps we'd need the ideal BE condensation formula, but again requires density.\n\nAlternatively
Rollout 23, Act: -2.411
min}^{-1}\n \]\n - Beta decay half-life \( T_{1/2,
Rollout 27, Act: -2.495
the Bose-Einstein condensation temperature? Because typical BEC occurs at nanokelvin to microkel
Rollout 23, Act: -2.522
decays into Y via alpha and into Q via Beta, but the activity isn't related to bateman
Rollout 27, Act: -2.570
_c related to when quantum effects become important? For BEC, T_c is the critical temperature below which
Rollout 23, Act: -2.623
decay and 0.098 minutes for beta decay. Let me make sure I understand what "
Rollout 23, Act: -2.657
2310 min−1.\n\nBut for beta decay:\n\nT_beta_1/2=0
Rollout 23, Act: -2.971
can decay in two different ways: alpha decay and beta-minus decay. The question asks for the activity of
Rollout 23, Act: -3.043
310 min−1\n\nSimilarly, for beta decay, the half-life is 0.0
Rollout 23, Act: -3.385
(Z+1,A) by decaying a $\beta^-$ with partial half life 0.0

Layer 17

GATE_PROJ

Top 16 Positive Activations
Rollout 54, Act: 3.158
For example, in an SN1 reaction, the RDS is the formation of carbocation. If that
Rollout 54, Act: 3.156
rance might make it less nucleophilic because the lone pairs are harder to access, but in solution,
Rollout 54, Act: 3.048
-stabilized anion; the charge is delocalized, which makes it a weaker nucleophile compared
Rollout 42, Act: 2.996
is typically 180°, for a bulky group in a strained environment.\n\nAssuming it's
Rollout 54, Act: 2.995
the following statements (B).\n\n1. In substitution reaction if the reactant in rate determining step is charged
Rollout 54, Act: 2.993
aren't an issue. However, with steric hindrance, bulky strong bases may have lower nucleophil
Rollout 44, Act: 2.991
1^gamma. So if gamma=5/3 for monatomic gas.\n\np1 = p2
Rollout 54, Act: 2.970
hydroxide?\n\nIn SN2 reactions, bulkiness hinders the backside attack, making it less
Rollout 54, Act: 2.941
hydroxide is better.\n\nBut with steric hindrance: 4-methylcyclohex
Rollout 54, Act: 2.941
protic solvents (from better to worse): RCOO− (carboxylate) <
Rollout 54, Act: 2.938
. Propionate is a carboxylate, RCOO−, so each of them except meth
Rollout 54, Act: 2.932
ophilic in water. Why? Maybe because the negative charge in hydroxide is more localized? So
Rollout 54, Act: 2.922
.\n\nSo even with steric issues, a localized charge (like 1) would make it more nucle
Rollout 44, Act: 2.921
ΔU = -W. For adiabatic expansion, W is positive (work done by gas
Rollout 47, Act: 2.902
is moving through a tunnel, and the cross-sectional area of the train is St, while the tunnel's
Rollout 54, Act: 2.901
HO− > RO−. This is because of polarizability and solvation. Thiolates (
Top 16 Negative Activations
Rollout 43, Act: -2.215
>= (alpha |0>+ beta |1> + gamma |1>) \tensorproduct |0>\n
Rollout 20, Act: -2.215
(an) �� 1 mod 43 and mod 47. To find the minimal such
Rollout 0, Act: -2.218
of 20! must go entirely to one of the numbers \(a\) or \(b\).
Rollout 21, Act: -2.223
that floor(x + n) = floor(x) +n for integer n. Maybe if we use that
Rollout 10, Act: -2.228
)\n\nBut since squares modulo4 are 0 or1. Therefore,n being0,1,2,
Rollout 21, Act: -2.233
=0 would imply that either f(x)=x or f(x)=x -1 for real numbers x
Rollout 10, Act: -2.253
subtract0 or1.\n\n subtract s≡0 or1 mod4, n -s≡0 -0
Rollout 21, Act: -2.281
for each integer k, f must take value k somewhere. Since functional equations are quantified over all variables
Rollout 20, Act: -2.283
prime p, either p ��1 mod m or p �� something else. So case 1
Rollout 85, Act: -2.299
0,0)$ is labeled by 0 , and for every $x, y \geq
Rollout 20, Act: -2.302
cases depending on whether p ��1 mod m or not.\n\nSo for modulus m (either 4
Rollout 43, Act: -2.360
\beta\left|1\right\rangle -\gamma\left|1\right\rangle
Rollout 63, Act: -2.362
1, q - 1) \) or \( (p - 1, q +
Rollout 64, Act: -2.367
=0. Then a_n= 0 +1*2^n=2^n. So it works
Rollout 21, Act: -2.438
f(x) such that either f(x)=x or f(x)=x-1, but those functions
Rollout 10, Act: -2.571
n ��0 mod4: subtract0 or1.\n\n subtract s≡0 or1 mod4,

UP_PROJ

Top 16 Positive Activations
Rollout 80, Act: 2.466
3)πr3 �� 4/3 * π * (1.2 × 1
Rollout 80, Act: 2.382
the volume of the inner core is (4/3)πr3. Let's calculate that:\n\n
Rollout 80, Act: 2.336
text{Volume} = \frac{4}{3} \pi r^3 = \frac{
Rollout 80, Act: 2.315
0^8 cm.\n\nVolume = (4/3)πr3 �� 4/3
Rollout 25, Act: 2.314
1 km, the volume would be (4/3)π*(3,541,0
Rollout 25, Act: 2.311
��, then mass m = (4/3)πR3ρ. So for R =
Rollout 81, Act: 2.282
resolve this?\n\nWait, tangents of angles and quadrants. If angle between two lines calculated by the
Rollout 26, Act: 2.238
, but let's check the calculations decimal-by-decimal.\n\nΔT=14.8 K.\n\n
Rollout 80, Act: 2.234
)π: \n\nVolume �� (4/3) * 3.1416 *
Rollout 99, Act: 2.229
x3, perhaps rounding to, say, four decimal places. Alternatively, keeping more decimals.\n\nAlternatively,
Rollout 80, Act: 2.210
pi r^3 = \frac{4}{3} \pi (1.2 \times
Rollout 80, Act: 2.203
Let's calculate that:\n\nVolume = (4/3)π(1.2 × 10
Rollout 69, Act: 2.169
0°=2/√3≈1.1547...\n\nTherefore, S= (1
Rollout 14, Act: 2.146
.31 which cubes to ~2.25, which is way small than8, indicating that
Rollout 99, Act: 2.141
perhaps during computation, we need to carry more decimal places.\n\nLet me redo the first iteration with more exact
Rollout 26, Act: 2.118
other values (Hg enthalpy and calorimeter capacity are given to two decimal places), likely the
Top 16 Negative Activations
Rollout 53, Act: -1.913
2 sqrt73 ) = (73 -27 ) / (2 sqrt73 )=
Rollout 93, Act: -1.915
2:\n\n(2*1007 -2j)/1007 = 2(
Rollout 52, Act: -1.926
2014*(2015 -2013)/2=2014
Rollout 38, Act: -1.927
552-2*1055 +1=1,113,02
Rollout 73, Act: -1.938
j > i} (2^j - 2^i). Our goal is to compute this sum
Rollout 67, Act: -1.938
3(n–1))*(-180 +24(n–1)). Let's check for n
Rollout 64, Act: -1.941
of (2^{2009} -2):\n\n= \( (a2)(2^{2
Rollout 18, Act: -1.948
( -1 -4 y), (4 -5 y). But since x divides all three differences,
Rollout 70, Act: -1.957
+ ... + d_1*(10 -2*7) + d_0*(1 -
Rollout 93, Act: -1.958
exponent:\n\nFirst, 2014 -2j: 2014=2*
Rollout 93, Act: -1.976
’}). But e^{i2pi m’}=1 if m’ is integer, so numerator becomes
Rollout 56, Act: -2.014
. For n≥2, 2^n -2 increases rapidly. Here, sequence of denominators:
Rollout 64, Act: -2.029
A is integer and k*2^{n -1} is integer for n≥1, a_n
Rollout 73, Act: -2.104
i, we have sum (2^j -2^i) = sum 2^j -
Rollout 73, Act: -2.104
^{j-1} (2^j -2^i). But our sum became N=1
Rollout 73, Act: -2.188
another way.\n\nEach difference 2^j -2^i for j > i. So maybe,

DOWN_PROJ

Top 16 Positive Activations
Rollout 98, Act: 2.275
ral type and mass: For example, a B0 star is around 20Msun, B
Rollout 48, Act: 2.238
hexadecimal value) in the IP header: 4500 0034 B61
Rollout 23, Act: 2.189
-2, A-4) by emitting an alpha particle with partial half life 3.0 minutes
Rollout 25, Act: 2.127
-body problem, the trajectory is determined by the gravitational parameter of the central body (Earth, in this case
Rollout 46, Act: 2.097
$U$, where $U=-\frac{G M_{1}}{\rho_{1}}-\
Rollout 98, Act: 2.092
help. Alternatively, I can recall that O, B, A are massive. Let me check: B
Rollout 23, Act: 2.081
Bateman equation if X decays into Y via alpha and into Q via Beta, but the activity isn
Rollout 25, Act: 2.073
- μ/(r), where μ is the gravitational parameter of the primary body (Earth, in this case
Rollout 98, Act: 2.071
WD (B5 when in the MS) and A0V\n\nG2V, K1V
Rollout 25, Act: 2.039
the specific angular momentum, and μ is the gravitational parameter (G*M_earth).\n\nBut if we consider
Rollout 48, Act: 2.034
- **Version and IHL:** `45` in hex. Version is 4 (IPv
Rollout 22, Act: 2.033
RT/(πM)), where R is the gas constant and M is the molar mass. Wait,
Rollout 25, Act: 2.029
6.674×10^-11 N m^2/kg^2, m is
Rollout 22, Act: 2.025
(π * M)).\n\nR is the gas constant, which is 8.3144
Rollout 87, Act: 2.011
of a cube where the cube has a uniform charge density rho, and each side is length a. Hmm
Rollout 3, Act: 2.002
(3RT/M), where R is the gas constant, T is the temperature in Kelvin, and M
Top 16 Negative Activations
Rollout 10, Act: -2.398
The key is to identify losing positions (P-positions) where every move leads to a winning position (
Rollout 42, Act: -2.408
carbocation. Wait, but if the mechanism is E1, then we first get carbocation formation.
Rollout 71, Act: -2.410
8041 -12d2 -12d. Now, 8041
Rollout 10, Act: -2.431
we need to find an infinite family of P-positions (Grundy number0). Since, even though
Rollout 42, Act: -2.449
, alcohols typically undergo elimination reactions, usually E1 or E2 mechanisms leading to the formation of
Rollout 10, Act: -2.461
n=2 is a losing position, so P-position.\n\nWait, contradicts my previous thought. Wait
Rollout 10, Act: -2.469
), positions can often be classified as either N-positions (next player can force a win) or P
Rollout 38, Act: -2.503
42857. Let's divide numerator and denominator by 48: 230
Rollout 59, Act: -2.523
6/784.\n\nFirst, divide numerator and denominator by 2: 123/
Rollout 90, Act: -2.540
. 112,000 -100,900=11,
Rollout 10, Act: -2.547
otic density zero, then the set of P-positions can still be infinite. The intuition being that when
Rollout 96, Act: -2.569
.\n\nWait, let me compute that. Divide numerator and denominator by common factors if possible. 17
Rollout 14, Act: -2.689
a -2)}.\n\nTake reciprocals on both sides:\n\nJ^{1/(2a -1
Rollout 10, Act: -2.695
next player can force a win) or P-positions (previous player can force a win). The key
Rollout 10, Act: -2.753
10, the losing positions (P-positions) are 0, 2, 5
Rollout 10, Act: -3.224
victory, those are called losing positions or P-positions in combinatorial game theory. The problem states

Layer 18

GATE_PROJ

Top 16 Positive Activations
Rollout 31, Act: 3.531
+1} + y a_n for n ≥ 0, where x and y are real numbers.
Rollout 68, Act: 3.056
has to be positive. Similarly, log(x + 2) is defined only when x + 2
Rollout 47, Act: 3.022
St, while the tunnel's is S0 = 4St. That means the area around the train
Rollout 32, Act: 3.010
let me recall that the roots of x3 = 1 are 1, ��, and �
Rollout 66, Act: 2.951
altitude is C1. Now, for n ≥ 2, Cn is the foot of the altitude
Rollout 81, Act: 2.922
the altitude divides BC into segments of 3 and 17, then the total length of BC is
Rollout 32, Act: 2.894
, since the sum of roots of x3 - 1 = 0 is zero (since the coefficient
Rollout 53, Act: 2.888
.5, PD = 2, DE = 2.5. Need to find the area of
Rollout 5, Act: 2.873
be b, then the longer base is b + 100. The midline length is (
Rollout 30, Act: 2.845
)(b_n + 1) for n ≥ 2,\n\nand\n\nb_{n+1}
Rollout 52, Act: 2.839
b, the sum is (b - a + 1)*(a + b)/2. So,
Rollout 65, Act: 2.821
a factor k, where 0 < k < 1. Similarly, if instead, the larger piece
Rollout 12, Act: 2.819
{dx}{\sqrt{(x^2 - 1)(1 - \nu^2 x^
Rollout 97, Act: 2.809
N2O5. Wait, but the problem doesn't specify the initial amount in moles. It
Rollout 44, Act: 2.805
Then irreversible work is 2*(20 -1)=38 L·atm. Reversible
Rollout 32, Act: 2.771
unity). So �� satisfies ��3 = 1 and �� ≠ 1. Also,
Top 16 Negative Activations
Rollout 98, Act: -1.316
coexistence.\n\n3. **G2V, K1V, M5V** \n -
Rollout 21, Act: -1.330
(z), floor applied to z in (-1,0) gives -1. So f(z) would
Rollout 98, Act: -1.333
0.9Msun star would be K-type).\n\n**Valid Systems**: 2 (WD +
Rollout 98, Act: -1.336
0V), 3 (G2V, K1V, M5V), 4 (
Rollout 98, Act: -1.349
existence.\n\n3. **G2V, K1V, M5V** \n - **
Rollout 98, Act: -1.408
on the MS would be a late G or early K star. The problem here is if it's called
Rollout 98, Act: -1.412
exist.\n\nThird system: G2V, K1V, M5V. All main-sequence
Rollout 21, Act: -1.427
- (n+1)∈ (-1,0). Hence z in (-1,0), Floor
Rollout 21, Act: -1.443
1,0). Hence z in (-1,0), Floor(z) is -1, so floor
Rollout 98, Act: -1.478
.8 Gyr). Additionally, an **A-type star** cannot have 0.9Msun
Rollout 98, Act: -1.542
and A0V\n\nG2V, K1V, M5V\n\nDA4, L
Rollout 21, Act: -1.547
- (n+1) ∈ (-1,0). And floor(z), floor applied to z in
Rollout 21, Act: -1.558
- (k+1)) = [-1,0). Applying f to that: ceil([-1,
Rollout 21, Act: -1.574
ceil(z). But for z ∈ (-1,0), ceil(z) =0. So ceil(x
Rollout 21, Act: -1.623
- (n+1) ∈ (-1,0). f(z)=ceil(z). But for z
Rollout 57, Act: -1.713
1). Therefore, C = (-infty,0] �� [1,infty). Now,

UP_PROJ

Top 16 Positive Activations
Rollout 20, Act: 2.298
903. 903 divided by 3 = 301. 3
Rollout 89, Act: 2.078
0/3. But 500 divided by 3 is about 166.6
Rollout 20, Act: 2.048
= 903. 903 divided by 3 = 301.
Rollout 70, Act: 1.990
147=6. Then 6 divided by 7=0, so digits should be
Rollout 48, Act: 1.926
25 rem 9. 225/16=14 rem 1.
Rollout 90, Act: 1.912
90. Is 1090 divisible by 2018? Let's check:
Rollout 90, Act: 1.905
426. 1426 divided by 2018: 201
Rollout 39, Act: 1.902
by 9. Since 999 divided by 9 is 111, 9
Rollout 89, Act: 1.883
. girls=66 (2/5 of 165=66), but boys=
Rollout 37, Act: 1.872
450. 450 divided by 2! (which is 2) is
Rollout 32, Act: 1.871
Number of cycles: 2009 divided by 3 gives 669 full cycles (
Rollout 48, Act: 1.866
5. 34,125 divided by 16 is 2132*
Rollout 92, Act: 1.865
517. 1517 divided by 17200? 17
Rollout 89, Act: 1.858
6 (2/5 of 165=66), but boys=99. Not
Rollout 0, Act: 1.851
2.\n\nWait, 7^2 divides into 20! but 7^3 does
Rollout 89, Act: 1.832
2/5. So 2/5 of 100 girls. But if girls are defined
Top 16 Negative Activations
Rollout 27, Act: -2.532
Alternatively, suppose that at T_c, the thermal energy k T_c is equal to the quantum energy per
Rollout 27, Act: -2.540
ms?\n\nAlternatively, maybe the problem assumes that the thermal energy at T_c is related to the kinetic energy
Rollout 22, Act: -2.547
.\n\nWait, I can refer to the formula via thermal de Broglie wavelength; for hydrogen.\n\nAlternatively
Rollout 37, Act: -2.553
is connected to the Newton’s identities, relating elementary symmetric functions to power sums.\n\nWait, another approach:
Rollout 27, Act: -2.554
Then, using the total rms speed to calculate the thermal temperature via (1/2)m V_Be
Rollout 27, Act: -2.574
problem might have typoed and swapped rms energy with thermal energy per particle at T_c. So ΔE
Rollout 27, Act: -2.577
)kT, which is half the average thermal energy per degree of freedom.\n\nWait, the problem says
Rollout 27, Act: -2.590
2)kT, which is half the average thermal energy per degree of freedom.\n\nWait, the problem
Rollout 4, Act: -2.591
are independent and random, and combines them in quadrature, while the max-min approach assumes that all errors
Rollout 37, Act: -2.593
any symmetric polynomial can be expressed in terms of elementary symmetric polynomials. The determinant is an alternating polynomial,
Rollout 27, Act: -2.622
Wait, what if the problem is mistaking the thermal temperature calculated here as T_c? If that's
Rollout 27, Act: -2.633
simplified and assumes that the critical temperature is where the thermal de Broglie wavelength is comparable to interparticle
Rollout 27, Act: -2.664
?\n\nAlternatively, maybe the problem assumes that the thermal energy at T_c is related to the kinetic energy via
Rollout 27, Act: -2.677
?\n\nAssuming that this V_Be is the thermal velocity at temperature T_c, then using the equip
Rollout 27, Act: -2.721
of sight velocity causing a Doppler shift in some energy levels? Wait, but the problem seems to want
Rollout 27, Act: -2.724
then estimate T_c. So perhaps it's about thermal motion causing some energy fluctuations, which relates to temperature

DOWN_PROJ

Top 16 Positive Activations
Rollout 66, Act: 2.769
altitude is C1. Now, for n ≥ 2, Cn is the foot of the altitude
Rollout 16, Act: 2.567
\n\nFirst, this problem reminds me of the coupon collector problem, where you have different coupons and you want
Rollout 54, Act: 2.412
CH2CH2COO−) is a resonance-stabilized anion; the charge is del
Rollout 54, Act: 2.389
a weaker base and less nucleophilic because of resonance stabilization?\n\nWait, propionate (CH2CH
Rollout 16, Act: 2.260
ally distributed variables isn't straightforward. In the coupon collector problem, each new coupon has a decreasing probability of
Rollout 16, Act: 2.258
expectation with indicators.\n\nRecall that for the coupon collector problem with n coupons, the expected time is n
Rollout 39, Act: 2.223
21} - 1)/(10 -1 ) = (10^{322}
Rollout 80, Act: 2.181
1 km = 1000 * 100 cm = 10^5
Rollout 90, Act: 2.174
be up to 10^{k} -1. But N must also be such that when multiplied
Rollout 77, Act: 2.167
4)=[14 / (14 -10)] * f(10,4)=
Rollout 77, Act: 2.165
And I need to compute f(14, 52). Hmm. Let's see.\n\nStarting
Rollout 16, Act: 2.119
each. The expected time should be similar to coupon collector. For coupon collector with n coupons, the expectation
Rollout 32, Act: 2.119
let me recall that the roots of x3 = 1 are 1, ��, and �
Rollout 2, Act: 2.101
)...(10^{999} -1). We need to compute P mod 10
Rollout 62, Act: 2.092
1) gives 11*(11 -1 +1)=11*11, which
Rollout 54, Act: 2.088
's more solvated. But propionate is resonance-stabilized → less nucleophilic. So
Top 16 Negative Activations
Rollout 93, Act: -1.707
, maybe we can approximate each term using Laplace's method or something similar for sums. For large ex
Rollout 99, Act: -1.711
.\n\nWait, perhaps another resource says:\n\nIn the Müller's method, to avoid inaccuracy from subtraction of
Rollout 12, Act: -1.723
Wait, let's recall that when we use differentiation under the integral sign, as long as the integral conver
Rollout 69, Act: -1.751
.16e-5. Cos(30/27≈1.111°
Rollout 51, Act: -1.752
So the absolute value can be split as follows:\n\n|sin(x - ��)| = {\nsin(x
Rollout 99, Act: -1.755
method works.\n\nFrom what I remember, in the Müller method, the next estimate x3 is found by
Rollout 93, Act: -1.775
equals �� [(expand into binomial, exchange orders, then some terms survive. )\n\nBut the sum
Rollout 99, Act: -1.781
step. So, we're supposed to use the Müller method to perform two iterations on the equation log1
Rollout 53, Act: -1.781
5\n\nSo, now we have three equations:\n\n1. (a + c)^2 + d^
Rollout 11, Act: -1.792
equations. But without real roots. But from Descarte’s signs, if the polynomial with real coefficients has
Rollout 16, Act: -1.792
/2)^{mk}\n\nThen, swapping the order of summation,\n\nE[T] = sum_{
Rollout 16, Act: -1.836
2)^{mk}\n\nThen, swapping the order of summation,\n\nE[T] = sum_{k
Rollout 99, Act: -1.839
, okay. \n\nFirst, I remember that the Müller method is a numerical method used to find the roots
Rollout 93, Act: -1.931
maybe we can approximate each term using Laplace's method or something similar for sums. For large exponents
Rollout 67, Act: -1.974
...(3)\n\nSo now we have three equations:\n\n1) A + B + C = 14
Rollout 87, Act: -2.016
x, then y, then z. Changing the order shouldn't matter because of the symmetry.\n\nBut integrating

Layer 19

GATE_PROJ

Top 16 Positive Activations
Rollout 80, Act: 1.608
text{Volume} = \frac{4}{3} \pi r^3 = \frac{
Rollout 80, Act: 1.565
0^8 cm.\n\nVolume = (4/3)πr3 �� 4/3
Rollout 80, Act: 1.498
the volume of the inner core is (4/3)πr3. Let's calculate that:\n\n
Rollout 25, Act: 1.334
1 km, the volume would be (4/3)π*(3,541,0
Rollout 80, Act: 1.193
4 cm3. Then multiply by (4/3)π: \n\nVolume �� (4/
Rollout 80, Act: 1.185
)π: \n\nVolume �� (4/3) * 3.1416 *
Rollout 32, Act: 1.184
$ be a nonreal root of $x^3 = 1,$ and let\n\[\math
Rollout 80, Act: 1.171
pi r^3 = \frac{4}{3} \pi (1.2 \times
Rollout 51, Act: 1.152
a helpful mathematics assistant.\n<|im_start|>user\nLet $\theta$ be a constant number such that
Rollout 80, Act: 1.146
3)πr3 �� 4/3 * π * (1.2 × 1
Rollout 56, Act: 1.125
) pasture. She starts with a triangle $A_{0} A_{1} A_{2}$
Rollout 80, Act: 1.124
24 cm3.\n\nMultiply by 4/3 π: 1.728 *
Rollout 22, Act: 1.116
02214076e23 molecules/mol):\n\n0.00201
Rollout 80, Act: 1.093
Let's calculate that:\n\nVolume = (4/3)π(1.2 × 10
Rollout 36, Act: 1.066
\n\nDivide all terms by 3:\n\ns4 - 30,624s2
Rollout 11, Act: 1.061
,b,$ and $c$ are three complex numbers with product $1$ .
Top 16 Negative Activations
Rollout 40, Act: -3.078
5π/2 divided by 2π is 75/4=18.75
Rollout 75, Act: -3.083
Wait, 32 divided by 4 is 8. So 8 can be written as sum
Rollout 20, Act: -3.085
021. Wait, but 1 multiplied by 1 is 1, so that would work
Rollout 75, Act: -3.101
1 + 3*0 = 3 mod 4. Alternatively, if we have seven squares (
Rollout 50, Act: -3.107
k\n\nWait, 20/2 is 10. So:\n\n11 = 1
Rollout 69, Act: -3.122
when i=0: 30° / 3^0 = 30°, right
Rollout 84, Act: -3.132
3/16 equals 2/9 of 27/32. So all checks out
Rollout 96, Act: -3.137
1740 divided by 13 is 1740 /13 is approximately
Rollout 93, Act: -3.153
2014 divided by 2 is 1007, and 100
Rollout 71, Act: -3.175
2011 �� 3 is 670.333... so not
Rollout 89, Act: -3.198
2/5. So 2/5 of 100 girls. But if girls are defined
Rollout 89, Act: -3.218
. girls=66 (2/5 of 165=66), but boys=
Rollout 89, Act: -3.229
Let's see, girls=2/5 of 150=60. boys=9
Rollout 71, Act: -3.303
8044 divided by 4 is 2011. So 3c2
Rollout 69, Act: -3.341
when i=0: 10° / 1 = 10°, and 3
Rollout 40, Act: -3.380
, 150 divided by 4 is 37.5, so 75π

UP_PROJ

Top 16 Positive Activations
Rollout 87, Act: 1.952
/2 and r from 0 to sqrt(2).\n\nThe problem then is splitting the integral into two
Rollout 12, Act: 1.835
t2) ) ~ 1 / sqrt(2ν(t - ν)(1 - ν2)
Rollout 12, Act: 1.735
integrand becomes ~ 1 / sqrt( 2ν u (1 - ν2) ), so
Rollout 12, Act: 1.705
the lower limit is ν, upper limit is1, and integrand depends on ν.\n\nSo by Le
Rollout 87, Act: 1.703
the first quadrant up to a radius of sqrt(2)*a). Wait, but integrating from y=
Rollout 87, Act: 1.629
square, r ranges from 0 to sqrt(2), and �� from 0 to π/
Rollout 9, Act: 1.608
, y). But the triangle being isosceles with base EM. Since triangle GEM is is
Rollout 5, Act: 1.605
(18125) = sqrt(25 * 725) = 5
Rollout 54, Act: 1.601
9-20) → weaker acid → stronger base.\n\n2. Hydroxide (pKa ~
Rollout 92, Act: 1.593
from eye), and the image is at the near point (63 cm from eye). However, need
Rollout 95, Act: 1.588
/3675^2)*sqrt(3675^2 +671*(
Rollout 87, Act: 1.581
integral equals 3 ln(1 + sqrt(2)) + sqrt(3) - 2.
Rollout 81, Act: 1.567
9 + h2) and AC=sqrt(289 + h2). Then angle at A
Rollout 54, Act: 1.559
polar protic solvent follows basicity due to charge density dominating. So more basic O-centered anions are
Rollout 81, Act: 1.553
0 (BC), b=AC= sqrt(289 +h2), c=AB=
Rollout 92, Act: 1.550
negative because it's a virtual image on the same side as the object). \n\nSo, plugging into
Top 16 Negative Activations
Rollout 51, Act: -2.013
}\n\nBut since our integral is over 0 to 2π, which can be partitioned into [
Rollout 5, Act: -2.015
segment connecting the midpoints of the legs divides the trapezoid into two regions with areas in the
Rollout 55, Act: -2.015
have this matrix A and I need to compute A raised to the 2018th power.
Rollout 56, Act: -2.017
starts with a triangle $A_{0} A_{1} A_{2}$ where angle $A
Rollout 49, Act: -2.017
statement is: If every subspace of a Hausdorff space X is σ-compact, then
Rollout 20, Act: -2.023
licative group modulo m. Wait, that is, kn must be a multiple of the exponent of the multip
Rollout 40, Act: -2.025
yi, then a rotation by π/4 counterclockwise around the origin would be multiplying by e
Rollout 20, Act: -2.043
of each prime power.\n\nThe problem wants σ(an) - 1 �� 0 mod 2
Rollout 46, Act: -2.065
:\n\nThe cross product term �� × v, in components assuming �� is along the z-axis:
Rollout 20, Act: -2.071
)^× is cyclic of order42, need kn ��0 mod42. But for kn
Rollout 10, Act: -2.082
\n\nWait, okay, so from 0 to 10, the losing positions (P-positions
Rollout 49, Act: -2.090
, a common example of an uncountable Hausdorff space that isn't σ-compact is
Rollout 2, Act: -2.095
remainder when the product of numbers like 9, 99, 999, and so
Rollout 65, Act: -2.099
. Hmmm.\n\nWait, perhaps we need that after cutting a prism by a plane parallel to one of
Rollout 20, Act: -2.113
But hold on, m=43: need kn �� 0 mod 42, and
Rollout 20, Act: -2.122
that exponent is 1.\n\nTherefore, we need kn �� 0 mod m - 1,

DOWN_PROJ

Top 16 Positive Activations
Rollout 49, Act: 3.958
show that if X is an uncountable Hausdorff space, then there exists some subspace
Rollout 32, Act: 3.939
let me recall that the roots of x3 = 1 are 1, ��, and �
Rollout 68, Act: 3.880
has to be positive. Similarly, log(x + 2) is defined only when x + 2
Rollout 96, Act: 3.703
perpendicular to L and contains point B(5, 6).\n- In the new coordinate system, line
Rollout 18, Act: 3.668
a pair (x, y) with x > 0 such that H is equal to H_{xy
Rollout 11, Act: 3.652
1/c = ab. So 1/a + 1/b + 1/c = bc + ac
Rollout 32, Act: 3.648
unity). So �� satisfies ��3 = 1 and �� ≠ 1. Also,
Rollout 49, Act: 3.499
, a common example of an uncountable Hausdorff space that isn't σ-compact is
Rollout 47, Act: 3.448
St, while the tunnel's is S0 = 4St. That means the area around the train
Rollout 69, Act: 3.429
the sine and cosine are 10° and 30° divided by 3^i,
Rollout 11, Act: 3.425
+ b + c) + (1/a + 1/b + 1/c), and q as
Rollout 33, Act: 3.394
,6}. For each k from 1 to 5, when we look at the first k elements
Rollout 66, Act: 3.392
altitude is C1. Now, for n ≥ 2, Cn is the foot of the altitude
Rollout 31, Act: 3.390
+1} + y a_n for n ≥ 0, where x and y are real numbers.
Rollout 30, Act: 3.386
-1} - 1)(b_n + 1). This relates a_{n+1}
Rollout 11, Act: 3.384
not sure. Alternatively, p is (a + 1/a) + (b + 1/b
Top 16 Negative Activations
Rollout 99, Act: -2.037
is2.1573. Hence, boxed answer as \boxed{2.157
Rollout 89, Act: -2.038
girls are in the class?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me see here. So the
Rollout 52, Act: -2.040
is the smallest yummy integer?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this problem about yummy
Rollout 25, Act: -2.048
$ is the impact parameter.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure this problem
Rollout 80, Act: -2.071
)π: \n\nVolume �� (4/3) * 3.1416 *
Rollout 80, Act: -2.084
3)πr3 �� 4/3 * π * (1.2 × 1
Rollout 80, Act: -2.105
pi r^3 = \frac{4}{3} \pi (1.2 \times
Rollout 83, Act: -2.113
the same if you can rotate one to get the other. So we have to consider rotational symmetry.\n\nSo
Rollout 25, Act: -2.210
00 m, then m = (4/3)π*(3.541e6
Rollout 98, Act: -2.221
\n(D) 2\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to figure out how
Rollout 80, Act: -2.377
Let's calculate that:\n\nVolume = (4/3)π(1.2 × 10
Rollout 80, Act: -2.390
0^8 cm.\n\nVolume = (4/3)πr3 �� 4/3
Rollout 25, Act: -2.407
��, then mass m = (4/3)πR3ρ. So for R =
Rollout 80, Act: -2.437
text{Volume} = \frac{4}{3} \pi r^3 = \frac{
Rollout 25, Act: -2.558
1 km, the volume would be (4/3)π*(3,541,0
Rollout 80, Act: -2.737
the volume of the inner core is (4/3)πr3. Let's calculate that:\n\n

Layer 20

GATE_PROJ

Top 16 Positive Activations
Rollout 85, Act: 2.545
all numbers are nonnegative integers and 0 is already at (0,0), can we have
Rollout 20, Act: 2.506
when you set k=1, it must hold, but then for higher k, it follows automatically.
Rollout 83, Act: 2.491
all arrangements have trivial stabilizer (no rotation fix any arrangement with unique digits, so every orbit indeed has
Rollout 95, Act: 2.447
3) k)/490 must equals rational number m →sqrt(3)=490*m
Rollout 10, Act: 2.439
number you add to S, you must exclude only finitely many others (i.e., n +
Rollout 35, Act: 2.391
4)=5. Since 5 and 1 is prime. Then solutions as factors would have same issues
Rollout 21, Act: 2.382
. But such f must satisfy for each integer k, f(k) =x where x is mapped under
Rollout 37, Act: 2.324
is divisible by (1-2)= -1, hence if m=1, this term is -
Rollout 70, Act: 2.316
to balance the equation, the higher digits must provide enough positive value to offset the negatives from the lower digits
Rollout 65, Act: 2.310
95, c. For similarity, this must satisfy that a1, 1995,
Rollout 64, Act: 2.307
negative, the term B*2^n needs to compensate it for all n.\n\nE.g., suppose A
Rollout 65, Act: 2.297
c2=1995*a must hold with a being <=1995 and c being
Rollout 64, Act: 2.285
-K y. Which for each y is exact, since both terms are integers. Thus, C +
Rollout 82, Act: 2.285
log2(2) (which is 1), but it's more complicated. Alternatively, if we
Rollout 21, Act: 2.275
) to be Z by making for every real y, f(y) must be in Z. But how
Rollout 8, Act: 2.264
are the only possibilities for mixed, and each is already a block of 1 red and 1 green
Top 16 Negative Activations
Rollout 40, Act: -2.549
3π/2} = cos(3π/2) + i sin(3π/
Rollout 69, Act: -2.550
phi\n\nBut not sure.\n\nAlternatively, write sin2(phi/3) = (1 - cos(
Rollout 93, Act: -2.554
π/1)]^{2}. So cos(π)^2 = (-1)^2 =1.
Rollout 93, Act: -2.556
) = -sinx. Therefore, cos(π/2 + π/(2*100
Rollout 3, Act: -2.561
. Let's check with calculator steps.\n\nAlternatively, sqrt(28,408). Let's
Rollout 93, Act: -2.580
1007}. And we know that cos2θ can be written as (1 + cos2
Rollout 69, Act: -2.583
handle.\n\nSo now, perhaps we can express sin2(theta) in terms that involve cos(3 theta
Rollout 17, Act: -2.600
e-11)^3 )\n\nThe 4πe-7 is μ0, which is
Rollout 69, Act: -2.612
trigonometric identities. Maybe we can write sin2(theta) using a double angle identity, or express
Rollout 93, Act: -2.615
07) and cos(1006π/1007) = -cos(
Rollout 93, Act: -2.642
=0.\n\nk=3: cos(3π/4)= -√2/2. same
Rollout 93, Act: -2.651
large even number, maybe I can express cos^{2014}(θ) in terms of multiple
Rollout 93, Act: -2.660
}^{1007} [cos(πk/1007)]^{20
Rollout 93, Act: -2.698
1006 would pair to cos(2π/1007) and cos(1
Rollout 55, Act: -2.727
and I need to compute A raised to the 2018th power. That seems like a
Rollout 93, Act: -2.738
2*1007). So cos(π/2 + x) = -sinx.

UP_PROJ

Top 16 Positive Activations
Rollout 48, Act: 2.085
the header contains error?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this IPv4 header in
Rollout 3, Act: 2.052
69 m/s.\n\n**Final Answer**\n\boxed{170} m/s\n\nWait no
Rollout 54, Act: 2.046
B = 2\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this question step by step
Rollout 3, Act: 2.045
3, so if need to answer, in the absence of the temperature in the problem, I should either
Rollout 3, Act: 2.044
.5 m/s. If using calculator accurate value.\n\nSo the answer is approximately 168.
Rollout 43, Act: 2.038
\right\rangle\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure out this problem
Rollout 3, Act: 2.030
69} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon gas is \boxed
Rollout 1, Act: 2.017
problem statement.\n\n**[Final Answer]**\n\boxed{\text{Such a point } y \text
Rollout 56, Act: 2.013
Therefore, we conclude.\n\n**Final Answer**\n\boxed{\sqrt{3}}\n\n<|im_start|>answer\nWes
Rollout 4, Act: 2.009
0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of the energy stored is
Rollout 64, Act: 2.006
is the desired answer.\n\n**Final Answer**\n\boxed{36 \left(2^{20
Rollout 99, Act: 2.005
of x_3?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through this problem
Rollout 29, Act: 2.002
Answer**\n\boxed{A}\n\n<|im_start|>answer\nGiven a gallon milk jug with a plugged handle due to
Rollout 22, Act: 1.999
But in several decimal places.\n\n\n**Final Answer**\nThe mean molecular speed of hydrogen gas (H2)
Rollout 25, Act: 1.989
text{ kg}}$\n\n\n**Final Answer**\nThe asteroid would need to be $\boxed{4.
Rollout 27, Act: 1.986
0^{6} \text{ K}}$.\n\n<|im_start|>answer\nGiven ΔE_rms =
Top 16 Negative Activations
Rollout 16, Act: -1.978
converted by day t is (1/2)^t. Therefore, the probability that all six saplings
Rollout 2, Act: -1.983
Thus, for k ≥3, 2^k �� 0 mod8, so (2
Rollout 51, Act: -1.997
x + x) + sin(x - 8x)] / 2 = [sin9x +
Rollout 72, Act: -2.028
root since n is even (since (-1)^n=1 when n even). Therefore, yes,
Rollout 2, Act: -2.029
=3. For k>=3:2^k mod8. But 2^3=8
Rollout 50, Act: -2.049
20x)^k = 20^k x^k \). So the equation is:\n\n
Rollout 73, Act: -2.064
sum 2^j - sum 2^i over all j >i.\n\nBut sum over all
Rollout 45, Act: -2.065
1 + (4 -m) - (5m +1) -4(m +1) ≥
Rollout 93, Act: -2.075
m)!)/(m!^2 *4^m}.\n\nWe also know that according to central bin
Rollout 45, Act: -2.075
-4) + (-4m -10m -4m)=18 -18m
Rollout 45, Act: -2.082
8 +16 -4m -10m -2 -4m -4= (8
Rollout 50, Act: -2.087
101 * (20/2)^k\n\nWait, 20/2 is
Rollout 2, Act: -2.121
\geq 3\), \(2^k \equiv 0 \mod 8\),
Rollout 61, Act: -2.227
9. Let’s write this as 2^m(1 + 2^{k - m})
Rollout 45, Act: -2.323
-1 -4) + (-m -5m -4m)=0 -10m ≥
Rollout 45, Act: -2.446
\n\nCalculate:\n\n1 +4 -m -5m -1 -4m -4= (1

DOWN_PROJ

Top 16 Positive Activations
Rollout 62, Act: 2.212
-generating polynomial? Maybe similar to Euler's famous polynomial n^2 +n +41, which
Rollout 34, Act: 2.142
out again:\n\n1. \( x(t) = 1 + \int_{0}^{t}
Rollout 62, Act: 2.017
's famous polynomial n^2 +n +41, which generates primes for n=0 to
Rollout 54, Act: 1.978
different anions: bulkier ones might be less nucleophilic because of steric hindrance. Also
Rollout 54, Act: 1.933
from the following statements (B).\n\n1. In substitution reaction if the reactant in rate determining step is
Rollout 54, Act: 1.932
more stabilized, but that's confusing) are better nucleophiles when considering polarizability. Or is
Rollout 54, Act: 1.913
actually, steric hindrance might make it less nucleophilic because the lone pairs are harder to access
Rollout 83, Act: 1.912
, here's a proper way to calculate:\n\nBurnside’s formula: count of orbits= (Sum over
Rollout 50, Act: 1.898
. Also, 22 is 2 × 11. Maybe these factorizations can help in
Rollout 54, Act: 1.897
aprotic solvents, alkoxide is more nucleophilic, but in polar protic, due
Rollout 54, Act: 1.886
and attack. Also, weaker bases might be poorer nucleophiles in polar protic solvents because stronger
Rollout 54, Act: 1.885
basicity. Steric hindrance also reduces nucleophilicity. The order is: \n\n**Ethan
Rollout 54, Act: 1.880
ions (e.g., sulfur-based) are better nucleophiles despite weaker basicity. Steric hind
Rollout 54, Act: 1.877
bases are more solvated, so their nucleophilicity is reduced. However, polarizability also
Rollout 69, Act: 1.873
the sine and cosine are 10° and 30° divided by 3^i,
Rollout 76, Act: 1.872
one might think the total number is 4 * 32 = 128. But wait
Top 16 Negative Activations
Rollout 30, Act: -1.666
_{n+1 }\n\nLet me check if this can telescope. From the recurrence relations. Let's recall
Rollout 27, Act: -1.684
value of the component of velocity along the line of sight).\n<|im_start|>assistant\n<|im_start|>think\nOkay, so
Rollout 29, Act: -1.696
's bottom. Thus, the answer would be A) P = rho g h. But according to options
Rollout 3, Act: -1.697
of some standard conditions. Hmm.\n\nWait, in kinetic theory, mean speed depends on temperature, so we
Rollout 70, Act: -1.701
7=3, remainder0; then 3 divided by7=0, remainder3. So digits
Rollout 72, Act: -1.704
onometric identities, we know that cos(72°) = 2 cos^2(3
Rollout 27, Act: -1.720
sight is one degree of freedom, then the average kinetic energy for that component is (1/2)
Rollout 69, Act: -1.728
(3 theta_i). Let's see if this can be manipulated.\n\nPerhaps use the identity for cos(
Rollout 27, Act: -1.728
?\n\nAlternatively, maybe it's the line of sight velocity causing a Doppler shift in some energy levels?
Rollout 47, Act: -1.745
$v_{t}$ inside a long tunnel of cross-sectional area $S_{0}\left(S_{
Rollout 27, Act: -1.758
. Alternatively, if the motion along the line of sight causes a Doppler shift in some emitted or absorbed
Rollout 27, Act: -1.761
to velocity. If the velocity along the line of sight is v_z, then the Doppler shift could
Rollout 89, Act: -1.794
number, then take 2/5 of that to get the number of girls. Let's confirm that
Rollout 27, Act: -1.822
the thermal energy at T_c is related to the kinetic energy via (3/2)kT_c
Rollout 95, Act: -1.860
the height of the triangle ABC with respect to base BC. Since area is 3√671
Rollout 27, Act: -1.896
1/2) kT as line of sight kinetic energy, then T = 2*ΔE

Layer 21

GATE_PROJ

Top 16 Positive Activations
Rollout 54, Act: 2.800
solution:\n\nIn SN2 reactions, the thiolate ion (e.g., SH−) is a
Rollout 54, Act: 2.767
stronger acids than alcohols, so thiolate is a weaker base? Wait, pKa of
Rollout 54, Act: 2.701
(CH3OH)\n5. Ethanethiolate (HS− or maybe C2H5S
Rollout 54, Act: 2.586
5S−? Wait, ethanethiolate would be CH3CH2S−)\n\nWait
Rollout 54, Act: 2.580
. Methanol, 5. Ethanethiolate) from most to poorest reactivity in aqueous
Rollout 54, Act: 2.576
? ~10, so ethanethiolate (conjugate base) is weaker base than
Rollout 54, Act: 2.537
even in polar protic solvents, thiolates (like SH−) are better nucleophiles
Rollout 54, Act: 2.519
in aprotic solvent, tert-butoxide would be better.\n\nSo, in polar protic
Rollout 54, Act: 2.514
(strong base)\n3. ethanethiolate? Wait, what's the pKa of eth
Rollout 54, Act: 2.514
3CH2SH, so ethanethiolate would be CH3CH2S−. Okay
Rollout 54, Act: 2.501
imolecular attack, so if the nucleophile is charged, solvation can hinder its reactivity. But
Rollout 54, Act: 2.487
same type of nucleophiles: oxygen-based anions. Between hydroxide (small, less polar
Rollout 54, Act: 2.484
ionate. That makes sense.\n\nEthanethiolate (HS− or RS−): Sulfur
Rollout 49, Act: 2.476
? Wait, no, σ-compact implies Lindelöf, but Lindelöf plus met
Rollout 54, Act: 2.468
:\n\n1. If reactant in RDS is charged (probably meaning the starting material is charged), and
Rollout 49, Act: 2.461
?\n\nNo. There are Hausdorff Lindelöf spaces that are not regular, called T
Top 16 Negative Activations
Rollout 64, Act: -3.097
2009} -1 )\n\nAlternatively, to solve for a2, let me write the inequality
Rollout 2, Act: -3.115
of1000 here?\n\nAlternatively, maybe because many of the terms in the product from k=
Rollout 23, Act: -3.130
0). No, mix of terms.\n\nAlternatively, think of Activity remains as initial of lambda_alpha *N
Rollout 93, Act: -3.132
m}.\n\n In our derivation process,mismatch arose due to the formula S= m*(2 + C
Rollout 65, Act: -3.153
x, but this seems not straightforward. Alternatively, think in terms of proportionality. If we have sides
Rollout 8, Act: -3.199
are adjacent and all greens are adjacent.\n\nAlternatively, think of the problem as the laid-out 5 cards
Rollout 69, Act: -3.200
and theta_{i+1}.\n\nAlternatively, think about possible telescoping products. But the sum is
Rollout 73, Act: -3.207
-k}. Wait, maybe not.\n\nAlternatively, think of it as sum_{i=0}^{
Rollout 96, Act: -3.216
to make sure the orientation is correct. Alternatively, maybe the angle for rotation should account for the direction.\n\n
Rollout 83, Act: -3.258
me cross-verify with another approach.\n\nAlternatively, think of arranging the residues in a cycle. Since each
Rollout 76, Act: -3.266
's a mistake in inclusion-exclusion.\n\nAlternatively, think of total colorings:512.\n\nNumber
Rollout 45, Act: -3.268
0, but let's see.\n\nAlternatively, think about setting x or y to 0. Let
Rollout 33, Act: -3.272
, but maybe this is not straightforward. Alternatively, think of assigning numbers step by step and making sure that
Rollout 5, Act: -3.287
but these are not similar necessarily. \n\nAlternatively, think algebraically. Let me denote the original trape
Rollout 23, Act: -3.291
0 min, not total decays.\n\nAlternatively, think of the problem using Bateman equation if X dec
Rollout 52, Act: -3.297
to be positive. Wait, but earlier reasoning was based on n and k from the formula where k is

UP_PROJ

Top 16 Positive Activations
Rollout 70, Act: 1.403
in base-10, D=2N, but N in 343 to 2
Rollout 42, Act: 1.366
in the cyclopropane, forming a more stable carbocation elsewhere. So if the cyclopro
Rollout 70, Act: 1.355
number D. So, D = 2N. Also, since the base-7 representation of N
Rollout 50, Act: 1.345
log_{2x}(202x)\). Using the change of base formula:\n\nLeft-hand side
Rollout 26, Act: 1.305
at 298 K. We need to find the standard molar enthalpy of formation of
Rollout 42, Act: 1.303
dride or alkyl shift to form a more stable carbocation. However, since cyclopropane
Rollout 43, Act: 1.296
1⟩) �� |0⟩. I need to apply the C-NOT gate to this input
Rollout 99, Act: 1.275
and x2 = 1. Then, we need to find the decimal value of x3. Hmm
Rollout 25, Act: 1.242
9898 km, and we need to find the required mass such that the perigee distance
Rollout 23, Act: 1.183
0385 s−1 *N(t).\n\nBut with N(t)=5e34 *
Rollout 49, Act: 1.177
σ-compact, then X must be countable. The conclusion is that X is countable as a
Rollout 49, Act: 1.169
σ-compact. Therefore, the contrapositive would be: If X is uncountable and Haus
Rollout 97, Act: 1.168
O4(g) and O2(g). The question says it's started in a closed cylinder under is
Rollout 23, Act: 1.165
ways: alpha decay and beta-minus decay. The question asks for the activity of the alpha decay after
Rollout 67, Act: 1.164
multiplying the corresponding terms of two arithmetic sequences. I need to find the eighth term of this sequence. Hmm
Rollout 95, Act: 1.161
B(9,0). Similarly, Qk is on AC, which goes from A(0,
Top 16 Negative Activations
Rollout 65, Act: -2.183
a must be a multiple of 1995. Since the original ordering has a ≤19
Rollout 65, Act: -2.186
2 /1995=1995. Therefore c is1995, but
Rollout 20, Act: -2.202
is42. Therefore, 2 is a primitive root modulo43.\n\nWait my prior thought was
Rollout 97, Act: -2.207
the rate can also be expressed in terms of partial pressure.\n\nThus, the rate of decomposition of N2
Rollout 20, Act: -2.212
6 respectively.\n\nIn cyclic groups, the multiplicative orders of elements are all the divisors of the group
Rollout 65, Act: -2.218
.\n\n 2. Cutting along 1995-axis, say y split such that smaller prism has
Rollout 93, Act: -2.227
om{2014}{2014} = 1\), we get:\n
Rollout 20, Act: -2.244
7)) )= LCM(1806, 2162). Which as before
Rollout 32, Act: -2.264
Key steps:\n\n1. Identify �� is a primitive cube root of unity, so ��3 =
Rollout 20, Act: -2.301
n must be a multiple of 1806 and 2162. So, n
Rollout 20, Act: -2.325
3*42/1=1806.\n\nSimilarly for m=47, lcm(
Rollout 20, Act: -2.353
multiple of the least common multiple of the multiplicative orders of all p modulo 43 and 4
Rollout 20, Act: -2.353
be the least common multiple of 1806 and 2162. Wait, but
Rollout 20, Act: -2.388
, meaning that the least common multiple of all possible orders (i.e., the exponent) is m -
Rollout 73, Act: -2.440
.\n\nYes, the formula is r/(1 - r)^2 - (n +1)r^{n
Rollout 20, Act: -2.446
, n must be LCM(1806, 2162). Let me compute

DOWN_PROJ

Top 16 Positive Activations
Rollout 57, Act: 1.958
First, regularity: A topological space is regular if, given a point x and a closed set
Rollout 98, Act: 1.901
ova. Wait, no. Stars above around 8 solar masses will go supernova and leave a neutron
Rollout 57, Act: 1.845
disjoint open sets separating x and C. Alternatively, regularity requires that closed sets and points can be separated
Rollout 49, Act: 1.838
is another line. X is hereditarily Lindelöf and T3 (regular and Hausd
Rollout 19, Act: 1.826
0.\n\nWait, let me do this step by step. Let's compute \( (k - 5
Rollout 42, Act: 1.818
the carbocation undergoes a Wagner-Meerwein rearrangement. If the initial carbocation is at
Rollout 69, Act: 1.806
here.\n\nAlternatively, since the problem involves an infinite product or sum maybe involving recursion, given the division by
Rollout 49, Act: 1.806
elöf.\n\nIn a hereditarily Lindelöf first-countable Hausdorff space
Rollout 49, Act: 1.792
ensation point.\n\nWait, in a Lindelöf space, every uncountable subset has a cond
Rollout 74, Act: 1.784
these.\n\nWait, let's approach this step by step. The compass needle points in the direction of the
Rollout 98, Act: 1.774
Wait, no. Stars above around 8 solar masses will go supernova and leave a neutron star or
Rollout 57, Act: 1.770
take the definition of regularity: a space is regular if and only if for every point x and closed
Rollout 49, Act: 1.766
use the fact that X is hereditarily Lindelöf and Hausdorff, hence it
Rollout 19, Act: 1.760
mod83. Let's compute this step by step.\n\nFirst compute 78 mod83=
Rollout 49, Act: 1.759
another way: since X is hereditarily Lindelöf and first-countable, let me assume
Rollout 49, Act: 1.754
editarily Lindelöf and T3 (regular and Hausdorff). To show that X
Top 16 Negative Activations
Rollout 72, Act: -1.652
⇒ Re(ζ)=0. Which would happen if �� is i or -i. But Re
Rollout 21, Act: -1.652
)=0 is set for all x, but that restricts f(x) ∈ {-1,0,
Rollout 57, Act: -1.654
. Csaszar*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that if we
Rollout 21, Act: -1.661
c. Then image is R, which doesn't fit. But perhaps adding another equation.\n\nAlternatively, perhaps
Rollout 82, Act: -1.669
9: 0,1\n\nBut in that case, codewords would be '70',
Rollout 37, Act: -1.684
$\det A$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I just came across this math problem
Rollout 15, Act: -1.693
off after some time?\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's try to figure out this problem
Rollout 97, Act: -1.724
~N}_{2} \mathrm{O}_{5}(g) \stackrel{\Delta}{\
Rollout 57, Act: -1.728
contained in the complement of U. That is possible if, for every y ∈ C, complement of U
Rollout 1, Act: -1.735
such sets have certain properties.\n\nAlternatively, I might need to use the parallelogram law. Let me
Rollout 21, Act: -1.737
) +1 -x=0, which only occurs at the boundaries. So that would hold only if
Rollout 49, Act: -1.752
is countable.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through this problem
Rollout 84, Act: -1.891
/9)*(1 - m/n). But that leads to a different equation.\n\nBut which interpretation is correct
Rollout 75, Act: -1.901
is 3 mod 4, which is possible if there are three squares that are 1 mod
Rollout 21, Act: -1.955
)) would also be f(x), but that only enforces that f is idempotent on its image
Rollout 21, Act: -1.960
image is a subset of Z. For that, maybe if we can write an equation that says f(x

Layer 22

GATE_PROJ

Top 16 Positive Activations
Rollout 1, Act: 2.851
$ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show
Rollout 69, Act: 2.816
Bonanza Team Round #10*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find
Rollout 58, Act: 2.813
Proposed by Yannick Yao*\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. So
Rollout 69, Act: 2.760
cos(30°/3^i)]\n\nI need to evaluate that infinite sum and express it as
Rollout 18, Act: 2.758
{(7, 5)}\).\n\n<|im_start|>answer\nTo solve the problem, we need to find a
Rollout 16, Act: 2.758
Proposed by Yannick Yao*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Emilia is trying
Rollout 0, Act: 2.744
**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and 1
Rollout 28, Act: 2.742
the centers of these hyperbolas.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I've got this
Rollout 45, Act: 2.735
possible value of $ m$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem
Rollout 70, Act: 2.730
51 = 102. Cool. I need to find the largest such N.\n\nFirst,
Rollout 2, Act: 2.717
109 in both cases. So The answer is109. Huh.\n\nBut wait,
Rollout 63, Act: 2.706
A1 for the three-variable case.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find
Rollout 28, Act: 2.697
, the midpoint of OG? Or related to medians.\n\nWait, in the general coordinate system, G
Rollout 31, Act: 2.675
{5}}{2}}\).\n\n<|im_start|>answer\nGiven real numbers \( x \) and \( y
Rollout 68, Act: 2.671
**\n\boxed{501}\n\n<|im_start|>answer\nTo find the number of integer values of \( k
Rollout 13, Act: 2.668
{187.12}\n\n<|im_start|>answer\nTo determine how much more money Joanie will owe
Top 16 Negative Activations
Rollout 3, Act: -1.249
,953.6; divided by π*0.222 �� 0.
Rollout 17, Act: -1.257
. So substituting:\n\nNumerator=2 *4πe-7 *1.41e
Rollout 3, Act: -1.261
as earlier for radon)\n\nDenominator: pi*0.028= ~0.0
Rollout 92, Act: -1.271
from lens, do = 20 cm -1.5 cm = 18.5 cm
Rollout 81, Act: -1.300
0)\n\n= numerator=1020 *121 =1020*12
Rollout 26, Act: -1.303
s.\n\nBut 20.00 *14.8: 14.8 has
Rollout 9, Act: -1.304
: y= (-5/25)x +50/25= (-1/5)x
Rollout 25, Act: -1.307
e20\n\nDivide by (2 *6.357e6 )=12
Rollout 3, Act: -1.308
.314*298/(π*0.222)) �� sqrt(
Rollout 56, Act: -1.310
, need exact fraction:\n\n1841 /1008 simplifies: Let’s see:\n\n
Rollout 3, Act: -1.317
.314*298/(π*0.028)) �� sqrt(
Rollout 3, Act: -1.322
811.04. Divide by π*0.222≈0.69
Rollout 36, Act: -1.334
,662,592 ��16=23,478,9
Rollout 16, Act: -1.356
=1/2. So term=6 *1 / (1/2)=6*2=
Rollout 17, Act: -1.391
with full numbers:\n\n|I|=(2 *4πe-7 *1.41e
Rollout 67, Act: -1.396
12,...\n\nMultiply them term-wise:\n\n8 *180=1440,\n\n11

UP_PROJ

Top 16 Positive Activations
Rollout 64, Act: 2.579
}^8 (M -Ky -0 +1 )? Wait, perhaps there's confusion here.\n\n
Rollout 46, Act: 2.497
} = -∇U - 2 �� × v + centrifugal term. Wait, no,
Rollout 25, Act: 2.480
m, via mu_final = G*(M + m). But if you need to achieve a smaller per
Rollout 64, Act: 2.468
solutions z is (M -K y ) +1= (2^{2012} -
Rollout 37, Act: 2.423
contribute a multiple of n! by orbit-stabilizer theorem. Therefore, the whole determinant is a sum
Rollout 53, Act: 2.415
Therefore, each median length: CE = CP + PE = 3.0 + 1.5
Rollout 64, Act: 2.394
_{y=0}^8 (M -Ky -0 +1 )? Wait, perhaps there
Rollout 28, Act: 2.368
3) -4*(k/3) +1=0\n\nExpand:\n\nMultiply through by9:\n\n
Rollout 28, Act: 2.344
a ) -4 (k /b ) +1=0.\n\nMultiply both sides by a2b
Rollout 71, Act: 2.335
= -3c2 + 8044, or equivalently, 3c2 +
Rollout 90, Act: 2.329
.\n\nBut let's start with 9N + M �� 0 mod 1009
Rollout 64, Act: 2.319
12} -1 -K y) +1. But 2^{2012}
Rollout 64, Act: 2.316
} -1 -K y -0 ) +1=2^{2012} -K
Rollout 52, Act: 2.307
(4028/3 -3 +1)/2= (1342.6
Rollout 16, Act: 2.306
/63. So 5 +11/217 -1 -1/63
Rollout 17, Act: 2.302
Tesla, **I** is the integral of **B** times probability density (unitless, since |
Top 16 Negative Activations
Rollout 78, Act: -1.343
. The answer is 49.\n\n**Final Answer**\n\boxed{49}\n\n<|im_start|>answer\n
Rollout 55, Act: -1.347
, the final answer is as above.\n\n**Final Answer**\n\boxed{\begin{pmatrix} \
Rollout 70, Act: -1.348
\boxed{315}.\n\n**Final Answer**\n\boxed{315}\n\n<|im_start|>answer
Rollout 58, Act: -1.357
037171}.\n\n**Final Answer**\n\boxed{203717
Rollout 38, Act: -1.370
is \boxed{61}。\n\n**Final Answer**\n\boxed{61}\n\n<|im_start|>answer\n
Rollout 53, Act: -1.374
frac{27}{2}}.\n\n**Final Answer**\n\boxed{\dfrac{27}{
Rollout 40, Act: -1.378
is \boxed{19}.\n\n**Final Answer**\n\boxed{19}\n\n<|im_start|>answer\n
Rollout 13, Act: -1.383
$187.12.\n\n**Final Answer**\n\boxed{187.12
Rollout 3, Act: -1.392
{169} m/s.\n\n**Final Answer**\nThe mean molecular speed of radon gas is
Rollout 25, Act: -1.395
} \text{ kg}}$\n\n\n**Final Answer**\nThe asteroid would need to be $\boxed{
Rollout 4, Act: -1.406
is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of the energy
Rollout 2, Act: -1.412
is \boxed{109}\n\n**Final Answer**\n\boxed{109}\n\n<|im_start|>answer
Rollout 97, Act: -1.418
boxed{2.30}。\n\n**Final Answer**\n\boxed{230}\n\n<|im_start|>answer
Rollout 73, Act: -1.477
, \boxed{398}\n\n**Final Answer**\n\boxed{398}\n\n<|im_start|>answer
Rollout 48, Act: -1.498
, the header does not contain errors.\n\n\n**Final Answer**\nThe IP header does not contain any errors.
Rollout 26, Act: -1.551
boxed{90.39}\n\n\n**Final Answer**\n\boxed{90.4}\n\n<|im_start|>

DOWN_PROJ

Top 16 Positive Activations
Rollout 93, Act: 2.102
1007 of [cos(πk/1007)]^{2014
Rollout 93, Act: 2.021
approach it. \n\nFirst off, exponents with cosine can sometimes be dealt with using power-reduction formulas
Rollout 66, Act: 2.012
**: The triangle \(ABC_0\) is a primitive Pythagorean triple, meaning its sides are of
Rollout 66, Act: 2.011
be pairwise coprime, it's likely a primitive triple where the legs are m2 - n2
Rollout 93, Act: 2.009
007. So [cos(πk/1007)]^{2*10
Rollout 66, Act: 1.971
is to find integers a, b, c (primitive Pythagorean triple) such that ab =
Rollout 66, Act: 1.967
=170\n\nCheck if this is a primitive Pythagorean triple: gcd(a, b )
Rollout 32, Act: 1.920
let me recall that the roots of x3 = 1 are 1, ��, and �
Rollout 45, Act: 1.915
possible value of $ m$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem
Rollout 7, Act: 1.914
71N.\n\nWhich simplifies as we did before to x=N/12.\n\nHowever, after
Rollout 66, Act: 1.909
be even. So anyway, the sides form a primitive triple.\n\nC1 is the foot of the altitude
Rollout 85, Act: 1.908
0,2024)$.\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to
Rollout 66, Act: 1.901
sides are pairwise coprime, it's a primitive Pythagorean triple. So a = m2
Rollout 30, Act: 1.896
n+3}}\right)$.\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I've got this
Rollout 7, Act: 1.887
-promoted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to
Rollout 93, Act: 1.859
1007} [cos(πk/1007)]^{2014
Top 16 Negative Activations
Rollout 18, Act: -1.727
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nConsider the additive group
Rollout 19, Act: -1.727
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the sum of
Rollout 31, Act: -1.727
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGiven are real numbers
Rollout 32, Act: -1.727
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $\omega$
Rollout 40, Act: -1.727
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA particle is located
Rollout 50, Act: -1.727
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nThere is a positive
Rollout 35, Act: -1.727
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $P(x
Rollout 61, Act: -1.727
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $S$
Rollout 59, Act: -1.727
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nJackie and Phil
Rollout 67, Act: -1.727
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind the eighth term
Rollout 58, Act: -1.727
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nPatchouli is
Rollout 88, Act: -1.731
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nConsider Convolutional
Rollout 4, Act: -1.738
1/C_eq = 1/C1 + 1/C2. So, I can write that as
Rollout 92, Act: -1.841
: 1/f = 1/do + 1/di. Here, do is the object distance
Rollout 4, Act: -1.845
/C_eq = 1/C1 + 1/C2. So, I can write that as C
Rollout 8, Act: -1.901
the number of sequences is the number of injective functions from 5 positions to 10 cards,

Layer 23

GATE_PROJ

Top 16 Positive Activations
Rollout 69, Act: 1.621
left side:\n\n=2√3 *a +2√3 *√b -3a -3
Rollout 96, Act: 1.608
132 + 12y) +25y =450\n\n12*
Rollout 36, Act: 1.604
*1,800*56 +562=3,240,0
Rollout 81, Act: 1.495
x2 +298x +9*289=x2 +298x +
Rollout 69, Act: 1.411
3 -3 ) /12 = (2*3 -3√3 ) /12√
Rollout 38, Act: 1.406
=(6*(24/7 x) +37x)/31=(144/
Rollout 28, Act: 1.404
)^2 = h2 + 2hk +k2 -4h -4k +4\n\n
Rollout 28, Act: 1.391
+k2 -4h -4k +4\n\nThen, expand (h -k)^2
Rollout 69, Act: 1.374
like terms:\n\n= (2a√3 +2√3 *√b ) -3a -
Rollout 4, Act: 1.370
in C_eq: sqrt( ( (10*3000^2)/(200
Rollout 36, Act: 1.367
Expand:\n\nx2 + x2 + (2 * x * 9000/s) +
Rollout 64, Act: 1.360
2009}-1 )\n\n= [8*2^{2009}+a1
Rollout 69, Act: 1.351
2 -√3 )√3 )/(4*3 )= (2√3 -3 )
Rollout 95, Act: 1.343
3(2k-1)/ (2*2450 )\n\n Distance between is (9 sqrt
Rollout 31, Act: 1.339
2]/4=(1 +2√5 +5)/4=(6 +2√5)/4
Rollout 4, Act: 1.329
5 -10*3000 +10*15=6,000
Top 16 Negative Activations
Rollout 56, Act: -3.140
in standard 30-60-90 triangles, the sides are 1 (opposite
Rollout 10, Act: -3.145
7-16=1, which is a winning position for opponent. So all moves from n=
Rollout 37, Act: -3.147
2. 3!=6, which does not divide 2. Hence, 2015
Rollout 20, Act: -3.148
) �� 1 mod 2021. Since this has to hold for all a,
Rollout 67, Act: -3.153
80 and 24. The GCD of 180 and 24 is
Rollout 15, Act: -3.156
]], which is identity. Therefore, not nilpotent. For n=3, as seen earlier,
Rollout 2, Act: -3.158
2^3 = 8 �� 0 mod 8, and 2^k for
Rollout 27, Act: -3.178
.491e5 m/s.\n\nYes, so approximately 1.491 ×
Rollout 62, Act: -3.200
*47= 517. Yes, 11*47. So two primes
Rollout 16, Act: -3.202
=1*2, so gcd is 1. Hence, 7880/19
Rollout 62, Act: -3.209
11*165.2 -> not. By 13:13*13
Rollout 37, Act: -3.217
. 3! =6, which does not divide 2. So n! is not a factor
Rollout 23, Act: -3.221
.51 Bq. Hmm, still closest to option B (3.719 Bq
Rollout 90, Act: -3.247
,396=448, 0? Let's compute 403,6
Rollout 96, Act: -3.261
3: 4+1=5, not divisible by 3. Hence no. So denominator isn
Rollout 81, Act: -3.299
^2=29584. Yes. So sqrt(29584)=1

UP_PROJ

Top 16 Positive Activations
Rollout 49, Act: 2.418
? Let me check.\n\nTake an uncountable discrete space D, and add a point at infinity,
Rollout 49, Act: 2.397
you take a subset that is uncountable and discrete, it would not be Lindelöf.\n\n
Rollout 49, Act: 2.389
countable and discrete. But an uncountable discrete space is not Lindelöf, so this
Rollout 49, Act: 2.369
the one-point compactification of an uncountable discrete space is not hereditarily Lindelöf
Rollout 49, Act: 2.314
able, maybe we can construct an uncountable discrete subspace?\n\nBut in hereditarily Lindel
Rollout 49, Act: 2.308
a subspace, which is uncountable and discrete. But an uncountable discrete space is not
Rollout 49, Act: 2.261
the one-point compactification of an uncountable discrete space is compact, hence Lindelöf,
Rollout 49, Act: 2.258
f, then it cannot contain an uncountable discrete subspace. But if X is uncountable
Rollout 49, Act: 2.249
the Cantor-Bendixson decomposition is a discrete subspace. For Lindelöf, all
Rollout 49, Act: 2.232
the one-point compactification of an uncountable discrete space is scattered but not hereditarily Lindel
Rollout 49, Act: 2.205
then, in a Lindelöf Hausdorff space, any closed set is also Lindel
Rollout 49, Act: 2.143
in this space, there are no uncountable discrete subspaces, so X is a continuous image of
Rollout 49, Act: 2.133
. Then A must contain a countable union of compact sets.\n\nEach compact subset in a Hausdor
Rollout 49, Act: 2.131
hereditarily Lindelöf spaces, every discrete subspace is countable. So if X is
Rollout 49, Act: 2.109
or-Bendixson decomposition is a discrete subspace. For Lindelöf, all discrete sub
Rollout 49, Act: 2.107
a hereditarily Lindelöf Hausdorff space is regular. Let's see: Take
Top 16 Negative Activations
Rollout 62, Act: -2.041
that expression is 11*q*q*p, total four primes.\n\nSo possible.\n\nSo testing this case
Rollout 56, Act: -2.045
√3 /144\n\nSum between brackets: (263√3 /144
Rollout 62, Act: -2.057
11*(13*31), so total three prime factors.\n\nSo for m=11
Rollout 62, Act: -2.062
1)=a*b*c, three primes. Then total prime factors would be 11*a*b*c
Rollout 75, Act: -2.063
2 + (2d)^2. Then total expression:4p +3= (2a)^
Rollout 62, Act: -2.092
12 gives 11*13 (total two primes).\n\nk=2:11*(
Rollout 89, Act: -2.104
66 and boys 100. Then fraction of girls is 66/166
Rollout 71, Act: -2.107
:202=400. then left 4022 -400=
Rollout 30, Act: -2.113
8,210.\n\nTherefore, the reduced form would be 4058211
Rollout 45, Act: -2.120
→ (n +c)x. Compare left which has n x. Therefore:\n\nn = n +c
Rollout 71, Act: -2.139
2=3969. Then remaining sum is4022 -3969=
Rollout 71, Act: -2.161
32=3969. Then remaining sum is4022 -3969
Rollout 70, Act: -2.161
57, so next digit 6. Finally remainder 48-6*7=6.
Rollout 62, Act: -2.196
*13*41. Three primes. Total three primes.\n\nStill insufficient.\n\nm=88
Rollout 45, Act: -2.207
x → (n +c)x. Compare left which has n x. Therefore:\n\nn = n +
Rollout 71, Act: -2.348
202=400. then left 4022 -400=3

DOWN_PROJ

Top 16 Positive Activations
Rollout 65, Act: 3.273
square, so primes in its factorization have even exponents. Therefore, since 1995
Rollout 65, Act: 2.691
is, if a2 has at minimum the exponents 1 in each prime, no matter how,
Rollout 0, Act: 2.664
k is primes. But, since prime factors with exponents: Wait, but each prime is either included
Rollout 65, Act: 2.515
at least 1 (since if a prime has exponent 1 in a, then exponent 2 in
Rollout 65, Act: 2.461
995 is square-free (all its prime exponents are1), then a must equals 1
Rollout 37, Act: 2.422
by all primes up to n but not their multiplicities. For example, 4 divides 2*
Rollout 52, Act: 2.381
divisors is (2+1)*(1+1)*(1+1)= 3*2*
Rollout 0, Act: 2.376
factorization of 20!.\n\nTo compute the exponents of primes in 20!:\n\nFor
Rollout 65, Act: 2.344
divides a2 only if in a2, the exponents for 3,5,7,1
Rollout 0, Act: 2.334
not a square (unless for example primes with even exponents), it depends. As with N! for
Rollout 65, Act: 2.325
situation is, if a2 has at minimum the exponents 1 in each prime, no matter how
Rollout 20, Act: 2.321
. To check:\n\nThe LCM takes the max exponents of each prime:\n\nPrimes involved:
Rollout 65, Act: 2.299
*5*7*19, each with exponent 1. So 1995 divides
Rollout 65, Act: 2.260
*7*19. Therefore, primes with exponent odd. Need "a" to be multiplied
Rollout 65, Act: 2.240
92. Number of divisors is (2+1)(2+1)(2+1)(
Rollout 37, Act: 2.206
015! is the product p1^e1 *p2^e2 *...*
Top 16 Negative Activations
Rollout 34, Act: -2.057
(s)\}ds$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have these two integral equations
Rollout 37, Act: -2.068
$\det A$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I just came across this math problem
Rollout 58, Act: -2.096
Yannick Yao*\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. So Patchouli
Rollout 93, Act: -2.099
007} for N=1008: No, wait, the angle was 2
Rollout 11, Act: -2.104
*David Altizio*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here where a
Rollout 16, Act: -2.107
Yannick Yao*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Emilia is trying to grow six
Rollout 49, Act: -2.113
is countable.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through this problem
Rollout 12, Act: -2.113
P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the function
Rollout 33, Act: -2.114
3,4),(3,5),(4,5).\n\nCompute each:\n\n(1,2):
Rollout 21, Act: -2.170
childkraut*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so part (a) is asking to
Rollout 37, Act: -2.193
any function f_x is periodic with period dividing p-1. Therefore, there are at most p-
Rollout 72, Act: -2.204
\n1. $ \n<|im_start|>assistant\n<|im_start|>think\nAlright, let's tackle this problem: Find all
Rollout 93, Act: -2.210
2^{2m}=4^{m}. Then C(2m,m)=4^m (sqrt
Rollout 69, Act: -2.263
Round #10*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find positive integers a
Rollout 28, Act: -2.361
these hyperbolas.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I've got this problem here about
Rollout 20, Act: -2.401
The exponent of a cyclic group is equal to the order of the group. Therefore, for a prime m

Layer 24

GATE_PROJ

Top 16 Positive Activations
Rollout 33, Act: 1.557
starts with 2 or3. Then, for k=2: first 2 elements must contain at
Rollout 20, Act: 1.447
is cyclic, there exist primitive roots where ord_m(p) = m -1, and elements of other
Rollout 33, Act: 1.432
, it can't be 1. Then for k=2, in the first two elements, there
Rollout 26, Act: 1.424
8 \mathrm{~K}$ and 1 atm into $\mathrm{HgO}(s)
Rollout 33, Act: 1.422
], [2,1]. The condition is for k=1: first element >1 (so invalid
Rollout 33, Act: 1.413
3 include numbers. However starting with4 will satisfy k=1 (not started with1) and k
Rollout 44, Act: 1.394
^(2/3). But T2= (p2 V2)/(nR). Wait, since
Rollout 22, Act: 1.374
73.15 K) and 1 atm pressure. Although pressure might not matter here because the
Rollout 20, Act: 1.364
necessary that kn is a multiple of the order of p modulo m. Since (Z/mZ)^×
Rollout 33, Act: 1.335
>2. Therefore, the only invalid permutations for k=2 are [2,1,*], i
Rollout 3, Act: 1.335
Pressure), which is 0°C and 1 atm. However, sometimes people use 25°C
Rollout 33, Act: 1.311
is 3. Therefore, the forbidden ones for k=2 are first two elements {1,2
Rollout 33, Act: 1.310
there's 5 choices here.\n\nThen, for k=2, the first two positions must not both
Rollout 20, Act: 1.302
residues mod m, we have that the order of p divides kn. Therefore, since the orders vary,
Rollout 33, Act: 1.289
is when the first element is 1. For k=2, the forbidden permutations are those starting with
Rollout 44, Act: 1.288
Because the gas does work against the highest possible opposing pressure. Conversely, for compression, the maximum work is
Top 16 Negative Activations
Rollout 65, Act: -3.404
proportion. Hmmm.\n\nWait, perhaps we need that after cutting a prism by a plane parallel to one
Rollout 62, Act: -3.414
multiple times. \n\nFirst, maybe I should start by understanding what the question is asking. If we have
Rollout 69, Act: -3.417
function.\n\nWait, another idea. Let me note that 3 theta_i = theta_{i -1
Rollout 50, Act: -3.420
maybe this isn't helpful. Alternatively, perhaps notice that 101 and 11 are primes
Rollout 11, Act: -3.420
Maybe considering algebra instead of analysis. Suppose we note that given p and q are real, can we relate
Rollout 24, Act: -3.426
relying on some underlying structure.\n\nPerhaps the key is that in a minimal saddle pair (R, C),
Rollout 50, Act: -3.441
log_{20x}(22x)\), which is equal to \(\log_{2x
Rollout 69, Act: -3.457
\n<|im_start|>think\nOkay, so I need to find positive integers a and b such that 1 over
Rollout 50, Act: -3.466
me recall some logarithmic identities that might help here. The change of base formula is useful for comparing logarith
Rollout 89, Act: -3.486
girls there are.\n\nSo the trick here is probably to figure out the total number of children first, right
Rollout 41, Act: -3.542
: For this problem, the user is likely expected to use the cases by number length and sum result (
Rollout 66, Act: -3.562
. So, maybe these lengths form a geometric series that sums to 6p. We have to find
Rollout 50, Act: -3.653
10}20\)\n\nLet me note that 202 is 2*10
Rollout 11, Act: -3.688
is 1. \n\nFirst, let me note that since abc = 1, maybe we can write
Rollout 30, Act: -3.707
term. Therefore, to telescope, we might need that -b_n + b_{n+2}
Rollout 89, Act: -3.838
is correct.\n\nWait, but maybe the key is that given boys are 100, and girls

UP_PROJ

Top 16 Positive Activations
Rollout 20, Act: 3.174
to have p^{kn} �� 1 mod m for all p not divisible by m, it
Rollout 20, Act: 3.134
we need p^{n} �� 1 mod 43. For k = 2,
Rollout 20, Act: 3.105
such that p^{n} �� 1 mod 43 and 47 for all primes
Rollout 20, Act: 3.067
3, p^{kn} �� 1 mod 43. Similarly, for modulus 4
Rollout 20, Act: 3.062
.\n\nTherefore, we need kn �� 0 mod m - 1, for m = 4
Rollout 20, Act: 3.013
reduces to p^{kn} �� 1 mod 43 for all primes p ≠ 4
Rollout 20, Act: 3.011
to have p^{n} �� 1 mod 43 for all p not 43
Rollout 20, Act: 2.976
thing is p^{kn} �� 1 mod m. Hmm. But how did k factor into
Rollout 20, Act: 2.968
p^{kn} \equiv 1 \pmod{43} \). This requires \( n
Rollout 20, Act: 2.968
we need p^{kn} �� 1 mod 47 for all primes p ≠ 4
Rollout 20, Act: 2.957
p^{kn} \equiv 1 \pmod{47} \). This requires \( n
Rollout 20, Act: 2.949
43, p^n �� 1 mod 43. Then, for any prime p
Rollout 61, Act: 2.947
2^{k - m}) �� 0 mod 9. Since 2 and 9 are
Rollout 20, Act: 2.941
Let's suppose that p^n �� 1 mod 43. Then, p^{kn}
Rollout 20, Act: 2.918
need p^{2n} �� 1 mod 43. However, but since if p
Rollout 20, Act: 2.905
=43: need kn �� 0 mod 42, and m=47:
Top 16 Negative Activations
Rollout 42, Act: -1.834
Here, the leaving group (water) has already left, forming a carbocation. Then, the base
Rollout 42, Act: -1.841
bond would form without rearrangement, but given the angle required for E2, which is typically 1
Rollout 31, Act: -1.842
6 -2√5)/4=(3 -√5)/2\n\nTherefore, y=2 -
Rollout 31, Act: -1.847
+√5)/2 and y=(1 -√5)/2:\n\nMultiply x and y:\n\n(
Rollout 29, Act: -1.857
looking back at the options, option E is P < rho g H. But according to the calculation,
Rollout 29, Act: -1.877
$ C) $\rho gH< P < \rho gh$ D) $P >
Rollout 42, Act: -1.915
not be very stable. Therefore, some carbocation rearrangement might occur. For example, maybe the cyc
Rollout 42, Act: -1.916
, and a hydrogen. Therefore, when the water leaves (after protonation), we need to eliminate a
Rollout 42, Act: -1.938
ring, it might lead to ring expansion or other rearrangements to relieve strain. Wait, especially the
Rollout 46, Act: -1.955
here, pointing along the z-axis if rotation is in the x-y plane), so the cross product �
Rollout 54, Act: -1.991
ic because steric hindrance would prevent it from approaching the electrophilic center as easily. So
Rollout 54, Act: -2.100
SN2 reactions, bulkiness hinders the backside attack, making it less effective. So maybe cyc
Rollout 29, Act: -2.136
gH\n\nC) rho gH< P < rho gh\n\nD) P > rho gh\n\n
Rollout 29, Act: -2.229
gH$ C) $\rho gH< P < \rho gh$ D) $
Rollout 54, Act: -2.370
In SN2 reactions, bulkiness hinders the backside attack, making it less effective. So maybe
Rollout 29, Act: -2.388
= rho gH\n\nC) rho gH< P < rho gh\n\nD) P > rho

DOWN_PROJ

Top 16 Positive Activations
Rollout 74, Act: 2.370
But in the problem statement, it just says "derive an expression for the deviation of the compass δθ
Rollout 20, Act: 2.348
m=43: need kn �� 0 mod 42, and m=47
Rollout 42, Act: 2.238
.\n\nWait, when the water leaves, the carbocation would form on the central carbon. However, having
Rollout 2, Act: 2.161
0 mod8\n\nThus, 5m≡0 mod8. So 5m is divisible by
Rollout 21, Act: 2.120
Schildkraut*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so part (a) is asking
Rollout 18, Act: 2.116
inverse of 3 modulo7, which is 5 (since 3*5=15≡
Rollout 20, Act: 2.113
1.\n\nTherefore, we need kn �� 0 mod m - 1, for m =
Rollout 20, Act: 2.110
p ��1 mod m, nk ��0 mod m. But here, wait. Wait,
Rollout 30, Act: 2.093
10.\n\nBut given the problem asks to "evaluate the infinite sum", likely as a fraction. But
Rollout 20, Act: 2.070
positive integer. But if we need kn ��0 mod m for all k, then n must �
Rollout 20, Act: 2.057
- 1. Therefore, kn �� 0 mod (m - 1) for each modulus
Rollout 56, Act: 2.050
at least visualize it. A0 is the right angle, A1 is 60°, A
Rollout 25, Act: 2.024
$ is the impact parameter.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure this problem
Rollout 20, Act: 2.022
if we take kn such that kn �� 0 mod (m - 1), which would be
Rollout 7, Act: 1.991
, but reading the problem again: "due to an error in the wording of a question, all scores
Rollout 57, Act: 1.975
*A. Csaszar*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that if
Top 16 Negative Activations
Rollout 50, Act: -2.199
which they are (since 101 is prime and 11 < 101 and
Rollout 49, Act: -2.225
(as above):\n\nLet X' = X without isolated points. If X' is non-empty, then
Rollout 50, Act: -2.227
that 101 and 11 are primes, and likely won't simplify. Perhaps I need
Rollout 14, Act: -2.228
2} dx is divergent at both 0 and ��. Hence this term is infinity. Thus
Rollout 19, Act: -2.239
or equal to6. Since 83 is prime, its divisors are 1 and83
Rollout 59, Act: -2.252
CD(246,784).\n\nPrime factors:\n\n246 = 2 *
Rollout 91, Act: -2.254
determined up to parametrization. Assuming D is origin, in our coordinate system. However, if we
Rollout 61, Act: -2.304
*13)* (2*5). So prime factors: 2^2 *3 *5
Rollout 62, Act: -2.369
13=> yes. 883 in prime? Yeah.\n\nSo m=99:1
Rollout 62, Act: -2.392
517=11*47. Prime.\n\nk=3: m=34.
Rollout 53, Act: -2.480
the centroid? Maybe coordinates system where centroid is the origin, or not. Let's try again.\n\nAlternative
Rollout 19, Act: -2.501
3 but >=6? 83 itself is prime, so only 1 and83. So
Rollout 20, Act: -2.559
2162? Let me check.\n\nPrime factors:\n\n1806: 2,
Rollout 76, Act: -2.762
=139. 139 is prime (since it's not divisible by 2,
Rollout 66, Act: -2.953
3 factors 13; 13 is prime, and 84 is 12*
Rollout 8, Act: -3.110
they can be simplified. Since 31 is prime. 126 divides by 2,

Layer 25

GATE_PROJ

Top 16 Positive Activations
Rollout 97, Act: 2.812
of N2O5}= P0 V / R T}= Assuming closed vessel.\n\nAt pressure P0
Rollout 75, Act: 2.674
32= 2? Wait hang on. Are you only even squares eligible?\n\nWait, consider
Rollout 97, Act: 2.447
partial pressure is proportional to concentration (PV = nRT, so n/V = P/(RT)). So
Rollout 3, Act: 2.399
(3RT/M), where R is the gas constant, T is the temperature in Kelvin, and M
Rollout 22, Act: 2.377
RT/(πM)), where R is the gas constant and M is the molar mass. Wait,
Rollout 45, Act: 2.350
that doesn't make sense. Wait, hold on. Let me correct.\n\nWait, wait, rhs should
Rollout 85, Act: 2.334
0)=4? Wait, starting to get confused.\n\nAlternatively, try to define the grids as 2
Rollout 22, Act: 2.325
3 K):\n\nv_mean = sqrt(8 * R * T / (π * M)) = sqrt
Rollout 22, Act: 2.269
by sqrt(8RT/(πM)), where R is the gas constant and M is the molar
Rollout 70, Act: 2.269
2 is in base-7 digits. Are.\n\nWait N=315:6 3
Rollout 10, Act: 2.266
1, which is winning. Hmm, wait. So if from n=5, can we move
Rollout 55, Act: 2.266
But actually, no, matrix multiplication is more involved. Wait, let's not jump the gun. Let
Rollout 22, Act: 2.247
3 K:\n\nv_mean = sqrt(8 * R * T / (π * M)), where\n
Rollout 22, Act: 2.246
8 K.\n\nv_mean = sqrt(8 * R * T / (π * M)).\n\nR
Rollout 44, Act: 2.244
pressure is not too far from the initial. Let me take p2=1.5 atm, V
Rollout 10, Act: 2.240
and odd. But maybe every other pair. Let me think. Alternatively, maybe in each two steps,
Top 16 Negative Activations
Rollout 90, Act: -3.783
891928}\n\n**Final Answer**\n\boxed{1008919
Rollout 81, Act: -3.783
510/11}\n\n**Final Answer**\n\boxed{\dfrac{510}{
Rollout 70, Act: -3.797
boxed{315}.\n\n**Final Answer**\n\boxed{315}\n\n<|im_start|>answer\n
Rollout 38, Act: -3.799
\boxed{61}。\n\n**Final Answer**\n\boxed{61}\n\n<|im_start|>answer\nAna
Rollout 93, Act: -3.806
clue. Given the process rigorously the Answer should be:\n\n(1007 / 2^{
Rollout 53, Act: -3.810
{27}{2}}.\n\n**Final Answer**\n\boxed{\dfrac{27}{2
Rollout 36, Act: -3.812
\boxed{192}\n\n**Final Answer**\n\boxed{192}\n\n<|im_start|>answer\n
Rollout 11, Act: -3.826
(-1, 3)}.\n\n**Final Answer**\n\boxed{(-1, 3)}\n\n<|im_start|>
Rollout 40, Act: -3.826
\boxed{19}.\n\n**Final Answer**\n\boxed{19}\n\n<|im_start|>answer\nThe
Rollout 52, Act: -3.851
{-2013}.\n\n**Final Answer**\n\boxed{-2013}\n\n<|im_start|>answer
Rollout 71, Act: -3.854
\boxed{98}.\n\n**Final Answer**\n\boxed{98}\n\n<|im_start|>answer\nGiven
Rollout 2, Act: -3.871
\boxed{109}\n\n**Final Answer**\n\boxed{109}\n\n<|im_start|>answer\n
Rollout 91, Act: -3.874
is \boxed{2}.\n\n**Final Answer**\n\boxed{2}\n\n<|im_start|>answer\nGiven triangle
Rollout 97, Act: -3.889
{2.30}。\n\n**Final Answer**\n\boxed{230}\n\n<|im_start|>answer\n
Rollout 25, Act: -3.894
\text{ kg}}$\n\n\n**Final Answer**\nThe asteroid would need to be $\boxed{4
Rollout 73, Act: -3.904
\boxed{398}\n\n**Final Answer**\n\boxed{398}\n\n<|im_start|>answer\n

UP_PROJ

Top 16 Positive Activations
Rollout 33, Act: 2.421
+ (-1)^m |A_1 �� A_2 �� ... �� A_
Rollout 79, Act: 2.404
*(Note: Wait, maybe "Changing Hams"? Not making sense. Wait apologies, after time-consuming
Rollout 64, Act: 2.390
is the desired answer.\n\n**Final Answer**\n\boxed{36 \left(2^{20
Rollout 23, Act: 2.371
might have an error.\n\n**Final Answer**\n\boxed{B}\n\n<|im_start|>answer\nTo determine the activity
Rollout 69, Act: 2.329
=204.\n\n**Final Answer**\n\boxed{204}\n\n<|im_start|>answer\nGiven the
Rollout 72, Act: 2.315
all even positive integers.\n\n**Final Answer**\n\boxed{\text{All even positive integers}}\n\n<|im_start|>answer
Rollout 65, Act: 2.303
. But subtracting\n\n**Final Answer**\n\boxed{40}\n\n<|im_start|>answer\nGiven a right
Rollout 76, Act: 2.296
3, wait 417: 4+1+7=12, so divisible by
Rollout 72, Act: 2.292
1 + 2Re(a)=1 implies Re(a)=0, which is not possible for roots
Rollout 77, Act: 2.279
4,52)= [52/(52-14)]*266 = (
Rollout 81, Act: 2.276
0/11}\n\n**Final Answer**\n\boxed{\dfrac{510}{11
Rollout 72, Act: 2.262
1 + 2Re(a)=1 ⇒ Re(a)=0. But Re(a)=0 implies
Rollout 11, Act: 2.261
4. But c = -1/4 is real, which is prohibited. Not allowed. So,
Rollout 84, Act: 2.259
5/14= (7/14 -5/14)=2/14=
Rollout 14, Act: 2.256
vectors / weighet\n\n**Final Answer**\n\boxed{ \left(\int_0^\infty
Rollout 72, Act: 2.252
144°)= cos(180° -36°)= -cos(36
Top 16 Negative Activations
Rollout 25, Act: -2.007
a_m = G*(M + m)/r2. Hence, in the frame of reference of one
Rollout 17, Act: -2.011
cos2θ - μ_p] divided by r3 multiplied by |Ψ(r)|2, then integrated
Rollout 97, Act: -2.015
atm of N2O5. For every 2 units of pressure (atm) of N2
Rollout 9, Act: -2.017
better way. Since each side is 10 units. So AI would be vertical, IM horizontal,
Rollout 96, Act: -2.026
passes through point A(24, -1). Let's confirm the equation for line L. The
Rollout 46, Act: -2.030
two terms -G M1 / ��12 * ��ρ1/∂x is the
Rollout 25, Act: -2.066
masses, F = G*M*m / d^2. Then, acceleration on the asteroid is F/m
Rollout 87, Act: -2.127
is sqrt(x2 + y2 + z2). Therefore, the integral becomes a triple integral over x
Rollout 25, Act: -2.156
asteroid is F = G*M*m / d^2, where M is Earth's mass, m is
Rollout 81, Act: -2.166
:\n\nAB= sqrt{32 +h2 }\n\nAC= sqrt{172 +h2
Rollout 25, Act: -2.194
acceleration due to gravity is G*M / d^2, solely dependent on Earth's mass.\n\nWait,
Rollout 38, Act: -2.196
+ W.\n\nBut wait, Ana has speed 8.6 m/s, so the time would be
Rollout 25, Act: -2.225
be accelerating with G*(M + m)/r2.\n\nTherefore, the relative motion can be modeled using
Rollout 25, Act: -2.227
relative acceleration is G*(M + m)/d2. Therefore, in the two-body problem, the
Rollout 97, Act: -2.227
rate of the reaction is given with rate constant 5 x 10^-4 s^-1.
Rollout 25, Act: -2.314
asteroid would be a = G*m / d^2, where d is the distance between them. Let

DOWN_PROJ

Top 16 Positive Activations
Rollout 10, Act: 2.289
can have subtractions only up to s <=n. For each n, it's a finite number
Rollout 90, Act: 2.279
k=2 digits. Wait no: hold on. Wait, for a 5-digit number: D
Rollout 79, Act: 2.229
" is a container, but I'm not sure.\n\nWell, "minor changes in" would be M
Rollout 90, Act: 2.216
2 digits. Wait no: hold on. Wait, for a 5-digit number: D1 D
Rollout 3, Act: 2.202
, temperature must be in Kelvin in the formulae.\n\nHence, given the confusion, perhaps the problem
Rollout 8, Act: 2.201
Wait, hold on. Wait, wait, no. If she is laying out 5 red cards,
Rollout 8, Act: 2.200
0. Wait, hold on. Wait, wait, no. If she is laying out 5 red
Rollout 33, Act: 2.195
\), at least one of the first \(k\) terms of the permutation is greater than \(k\
Rollout 0, Act: 2.192
to their exponents).. Wait, confusion arises here.\n\nWait, actually, we need to be precise.
Rollout 44, Act: 2.184
process to be both adiabatic and isothermal, Q=0 and ΔU=0 (
Rollout 10, Act: 2.140
s) | s is a square, s<=n}.\n\nSo let's compute the Grundy numbers
Rollout 20, Act: 2.136
1. Wait, no—Wait, wait.\n\nWait, no. Let's suppose that p^n
Rollout 8, Act: 2.125
10. Wait, hold on. Wait, wait, no. If she is laying out
Rollout 44, Act: 2.110
to V2.\n\nWork done by gas in isothermal reversible expansion: W_iso = nRT ln(V
Rollout 44, Act: 2.107
than irreversible.\n\nSo in that case, since the reversible work is larger, then [A] would be
Rollout 26, Act: 2.094
so that’s why T=298 K.\n\nThus, in the calculation for ΔH, the
Top 16 Negative Activations
Rollout 17, Act: -2.451
π a03 )\n\nBut μ0 has units T·m/A, μ_p has units J/T
Rollout 59, Act: -2.465
16] / 282\n\nCompute numerator:\n\n9 + 100 = 1
Rollout 17, Act: -2.477
/A * J/T ) / m3 = (J·m) / (A·m3)
Rollout 96, Act: -2.486
/ (169*13)\n\nCompute numerator:\n\n-4106*12=
Rollout 17, Act: -2.571
) = kg/(A·s2) = T, since 1 T = kg/(A·
Rollout 97, Act: -2.582
2y)\n\nBut solving this, we can rearrange variables:\n\ndy/(1 -2y)
Rollout 3, Act: -2.593
8*R*T)/(π*M)), R is in J/(mol·K), which is same as (
Rollout 17, Act: -2.597
* J/T ?\n\nWait, μ0 has units T·m/A, μ_p has units J/T
Rollout 97, Act: -2.666
rearrange variables:\n\ndy/(1 -2y) = (k / 2) dt\n\nInt
Rollout 97, Act: -2.682
solving this, we can rearrange variables:\n\ndy/(1 -2y) = (k /
Rollout 80, Act: -2.696
So, the volume of the inner core is (4/3)πr3. Let's calculate
Rollout 17, Act: -2.758
) = kg/(A·s2) = T, which matches. Correct. So the result is
Rollout 80, Act: -2.872
the volume of the inner core is (4/3)πr3. Let's calculate that:\n\n
Rollout 17, Act: -2.889
·m/A * J/T \n\nBut J = N·m = (kg·m2/s2
Rollout 76, Act: -2.903
entirely red squares. It might be easier to compute the number of colorings that have at least one
Rollout 97, Act: -3.099
ange variables:\n\ndy/(1 -2y) = (k / 2) dt\n\nIntegrate

Layer 26

GATE_PROJ

Top 16 Positive Activations
Rollout 20, Act: 3.878
's see, 43*900=38,700; 43
Rollout 96, Act: 3.871
1 Check:\n\n13*8000=104,000\n\n13
Rollout 96, Act: 3.809
3: 13*6000=78,000, 88
Rollout 81, Act: 3.793
*561:\n\n- Calculate 40*561=22,440
Rollout 96, Act: 3.755
13: 13*649=843, and my previous step had
Rollout 53, Act: 3.723
146:\n\n146*13=1898 → 1971
Rollout 96, Act: 3.717
compute:\n\n4106*100=410600\n\n410
Rollout 36, Act: 3.673
624:\n\n30,000*30,000=900
Rollout 20, Act: 3.643
:\n\nFirst compute 40,000*47=1,880,0
Rollout 36, Act: 3.636
let's check:\n\n10,000*10,000=100
Rollout 90, Act: 3.608
: 2,000*44=88,000; 18
Rollout 17, Act: 3.590
400*3.1416=1256.64\n\n44
Rollout 96, Act: 3.565
37 divisible by13: 13*649=843, and my
Rollout 36, Act: 3.557
064: 8,000*8,000 = 64,
Rollout 96, Act: 3.553
. Wait 13*2000=26,000, remaining 2
Rollout 62, Act: 3.553
? Calculating 23*100=2300, 23*7
Top 16 Negative Activations
Rollout 28, Act: -3.215
to non-degenerate hyperbolas. If so, then the statement would require that all non-degenerate
Rollout 43, Act: -3.223
is first, then target qubit.\n\nIn that case, the input is (control qubit)⊗
Rollout 11, Act: -3.226
to both p and q being real.\n\nHence, maybe such a case is impossible, so perhaps we
Rollout 66, Act: -3.227
solution exists under this approach, but the problem states that such a triangle exists.\n\nWait, perhaps I made
Rollout 38, Act: -3.230
can't be right.\n\nHence, there's a flaw in the assumption that the path for Ana is
Rollout 35, Act: -3.246
8. \n\nBut wait: But the problem states that the equation P(n)=n +3 has two
Rollout 29, Act: -3.276
milk and handle's milk are separate.\n\nIn that case, the pressure at the bottom of the main jug
Rollout 66, Act: -3.282
be both odd. Therefore, the error arises here. Therefore, m and n must have opposite parity.
Rollout 0, Act: -3.284
, numerator a and denominator b must satisfy that. However, a different way to phrase this is: a
Rollout 50, Act: -3.304
)\) is indeed the correct value. Thus, the final answer is \(11 + 10
Rollout 64, Act: -3.304
, but the problem states that the sequence is of positive integers. So all terms a_n must be positive
Rollout 66, Act: -3.312
≠1. Thus, problem.\n\nHence, the approach is leading to non primitive triples. Which suggests
Rollout 20, Act: -3.353
any positive integer a, including non-primes. However, since σ is multiplicative, which allows us
Rollout 95, Act: -3.365
being trapezoids except the first triangle. Therefore, if each trapezoid has the same
Rollout 31, Act: -3.376
m. Wait a minute, did we maybe overlook the role of m?\n\nHmm, that may be a
Rollout 84, Act: -3.440
.5, so that's impossible. Thus, the only possible valid interpretation is the first one. Therefore

UP_PROJ

Top 16 Positive Activations
Rollout 84, Act: 1.649
2} - 2 \times \frac{m}{n} \).\n\nAccording to the problem,
Rollout 83, Act: 1.552
the number of distinct arrangements is 1296 / 9 =144.\n\nBut let
Rollout 83, Act: 1.529
}{9} \left( 1296 + 0 + 0 + \cdots
Rollout 36, Act: 1.518
$BP=60\sqrt{10}$ , $CP=60\sqrt{5
Rollout 36, Act: 1.505
sqrt{10}$ , $CP=60\sqrt{5}$ , $DP=1
Rollout 80, Act: 1.471
1} \, \text{neutrinos/s/cm}^2 \times 4.5
Rollout 79, Act: 1.470
it's phrased is "Noticed minor changes in investigators' facility," making "minor changes in
Rollout 83, Act: 1.445
. Hence, number of orbits is1296/9=144.\n\nThus, answering
Rollout 9, Act: 1.431
triangle GEM and square AIME is 80. Since the square is from (0,0
Rollout 27, Act: 1.431
54 × 10^-17 J, calculate the rms speed of the Be nuclei, V
Rollout 5, Act: 1.423
(H - h) = 62.5H.\n\nSo we have two equations:\n\n1.
Rollout 66, Act: 1.420
sum of this series is given as \(6p\). The key equation derived from the geometric series sum
Rollout 95, Act: 1.412
ABC [ (k/2450)^2 - ( (k-1)/245
Rollout 5, Act: 1.407
x)/2 * h = 62.5H. The remaining trapezoid has bases
Rollout 9, Act: 1.407
\) and square \(AIME\) is 80 square units. We need to find the length of
Rollout 53, Act: 1.403
:\n - Using \(DE = 2.5\), we get \((a + c)^
Top 16 Negative Activations
Rollout 74, Act: -2.986
, and $\theta$.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to derive an expression
Rollout 92, Act: -2.992
power of his glasses lenses?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure this out
Rollout 22, Act: -2.997
2) in m/s\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to calculate the mean
Rollout 46, Act: -2.999
^{2}\right)$.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to find the second
Rollout 31, Act: -3.007
2: x^3 + x^2. Subtract from dividend:\n\n(x^3 -0x^
Rollout 3, Act: -3.011
n) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular
Rollout 17, Act: -3.014
$ (in SI units).\n<|im_start|>assistant\n<|im_start|>think\nAlright, let me try to tackle this problem
Rollout 97, Act: -3.016
$\mathrm{Y}$?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I have this
Rollout 62, Act: -3.022
. Solve.\n\nTry m values:\n\nm=0:-2≡-2≠0\n\nm=1
Rollout 29, Act: -3.024
\rho gH$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure out this
Rollout 25, Act: -3.025
is the impact parameter.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure this problem out
Rollout 47, Act: -3.069
value of $N$?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so there's this problem about a
Rollout 26, Act: -3.079
mol}^{-1}$ ]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's try to figure this problem
Rollout 80, Act: -3.110
value of $N$ ?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure out this
Rollout 25, Act: -3.133
$ is the impact parameter.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure this problem
Rollout 27, Act: -3.167
along the line of sight).\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I've got this problem here

DOWN_PROJ

Top 16 Positive Activations
Rollout 22, Act: 2.819
333e3 *1e3≈ (19.820576 /
Rollout 22, Act: 2.669
6.333e-3 �� (18,168.776
Rollout 54, Act: 2.633
ophiles given are:\n\n1. 4-methylcyclohexan-1-olate\n
Rollout 4, Act: 2.630
990 * 2985= (2000 -10)*(30
Rollout 54, Act: 2.530
But with steric hindrance: 4-methylcyclohexan-1-olate is
Rollout 22, Act: 2.530
do 66 * 273 = (60*273) + (6
Rollout 36, Act: 2.518
?\n\n19,360^2= (19,000 + 36
Rollout 38, Act: 2.496
, 196/672= (28*7)/(28*24
Rollout 13, Act: 2.487
09\n\n1.03^4 = (1.0609)^2\n\nCompute
Rollout 70, Act: 2.478
2*51=102. Correct, which is the example given.\n\nd1=1
Rollout 48, Act: 2.447
00 + 0xB612 = (0x4500 + 0xB
Rollout 22, Act: 2.416
0.006333\n\n= (18,157.776
Rollout 54, Act: 2.375
than water. Wait, but comparing 4-methylcyclohexanol's pKa—probably
Rollout 3, Act: 2.372
3 * 66.512 = (200 +70 +3)*6
Rollout 39, Act: 2.349
is 1 followed by 322 zeros minus 10. So the result will be a
Rollout 69, Act: 2.317
3 -3 )/12. That's correct.\n\nTherefore, S= (2√3 -
Top 16 Negative Activations
Rollout 60, Act: -1.967
the sound at the frequency it has received).\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to
Rollout 38, Act: -1.973
sides are the north, south edges; the shorter sides are east and west edges. Wait, that seems
Rollout 76, Act: -1.993
What's the intersection of S1, S2, and S3? Each pair of these overlaps:\n\n
Rollout 76, Act: -2.027
two 2x2 squares are both all red. How many such pairs are there, and what's
Rollout 15, Act: -2.030
i-1,t} + x_{i+1,t} mod 2\n\nThis is similar to
Rollout 16, Act: -2.047
8). 1953=68*28 + 49\n\n68=
Rollout 42, Act: -2.065
utyl or cyclopropyl side? Or wait, maybe the adjacent carbons are not on the
Rollout 87, Act: -2.072
0}$, and any necessary numerical constants.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to
Rollout 33, Act: -2.113
, according to formula for n=4:\n\nUsing inclusion-exclusion:\n\n24 -16( m
Rollout 22, Act: -2.142
hydrogen (H2) in m/s\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to
Rollout 83, Act: -2.148
of residues 0,1,2 to Group0,1,2), the number of distinct label
Rollout 76, Act: -2.177
are both all red. How many such pairs are there, and what's the number of colorings for
Rollout 74, Act: -2.182
_{s}$, and $\theta$.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to
Rollout 76, Act: -2.188
using the inclusion-exclusion principle.\n\nSo, apply inclusion-exclusion. Let the four 2x2
Rollout 3, Act: -2.205
on (Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate
Rollout 76, Act: -2.268
we need to adjust for overcounting using the inclusion-exclusion principle.\n\nSo, apply inclusion-exclusion

Layer 27

GATE_PROJ

Top 16 Positive Activations
Rollout 97, Act: 2.872
?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I have this decomposition reaction: 2
Rollout 23, Act: 2.839
\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this problem step by step. So, we
Rollout 26, Act: 2.819
]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's try to figure this problem out. So, we
Rollout 91, Act: 2.794
\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this geometry problem step by step. The question
Rollout 5, Act: 2.782
.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve this problem about a
Rollout 65, Act: 2.760
?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see, I've got this problem here about a
Rollout 64, Act: 2.738
?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's try to figure out this problem. We have a
Rollout 54, Act: 2.718
\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this question step by step. First, I
Rollout 52, Act: 2.695
?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this problem about yummy integers. The question is
Rollout 8, Act: 2.689
.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need to find the probability that when
Rollout 58, Act: 2.669
*\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. So Patchouli has an exam with
Rollout 67, Act: 2.636
.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. The problem says that the sequence starts with
Rollout 40, Act: 2.623
.\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's tackle this problem step by step. So, we
Rollout 62, Act: 2.613
.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this problem: Find the least positive integer \(
Rollout 95, Act: 2.582
.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve this geometry problem about
Rollout 31, Act: 2.570
.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's try to tackle this problem. So, we have
Top 16 Negative Activations
Rollout 93, Act: -3.484
6)/16=2*8/16=1). For m=3:3*(
Rollout 0, Act: -3.493
128. Alternatively, total number of subsets is 2^8=256, each
Rollout 93, Act: -3.512
formula gives sqrt(3/3.14≈0.97 whereas actual formula gives33
Rollout 64, Act: -3.518
this.\n\nBut 36=4*9, so may be written as9*4*(2
Rollout 69, Act: -3.521
Alternatively, 4 - ��3 ��4 -1.732≈2.2
Rollout 39, Act: -3.521
9, total is 318*1 +789, but how exactly?\n\nAlternatively,
Rollout 73, Act: -3.529
} +2=9*2048 +2=18432 +2=
Rollout 86, Act: -3.547
/8. Numerical value: pi^3≈31.006, so 3
Rollout 33, Act: -3.568
.\n\nBut wait 6! equals720 , the answer is 461 which makes sense
Rollout 0, Act: -3.611
4, which is 2^{3-1}=4.\n\nSimilarly, if n is a square-free
Rollout 70, Act: -3.639
? Wait, wait N=35*9 +16=315 +16=
Rollout 81, Act: -3.653
39270= divide numerator and denominator by 10: 12342
Rollout 51, Act: -3.712
2/7= (14 -18)/63 = (-4)/63.\n\nTherefore
Rollout 0, Act: -3.745
count above). But 2^{3-1}= 4 correct. Hence, Therefore, the formula
Rollout 75, Act: -3.811
So then scaling back, sqrt(4×4)=4* 2^2. This is similar
Rollout 86, Act: -3.812
pi^3≈31.006, so 31.006/4

UP_PROJ

Top 16 Positive Activations
Rollout 2, Act: 2.337
8, so 10^k mod 8 is 2^k mod 8.\n\nThus
Rollout 19, Act: 2.292
But also, note that congruence modulus k. So need x2 ��17 modk
Rollout 19, Act: 2.252
.\n\nBut also, note that congruence modulus k. So need x2 ��17 mod
Rollout 2, Act: 2.243
(10^k -1) mod 8 is (2^k -1) mod
Rollout 2, Act: 2.229
Since 10 �� 2 mod 8, so 10^k mod 8
Rollout 2, Act: 2.216
^k -1) �� -1 mod8.\n\nTherefore, for k=1: (2
Rollout 81, Act: 2.201
Let me calculate angle BAC by vectors. AB vector from A to B: (-3, -h
Rollout 71, Act: 2.185
+ m, where m= -abc. Okay.\n\nWe need integers a, b, c with a
Rollout 73, Act: 2.181
4,8,...,1024}. So for each pair with larger element 2^j
Rollout 21, Act: 2.144
?\n\nWait, suppose f: R→Z. Then, f(f(x) + y) is equal
Rollout 81, Act: 2.140
AC by vectors. AB vector from A to B: (-3, -h). AC vector from A
Rollout 2, Act: 2.140
8.\n\nCompute 2^k mod 8:\n\nNotice that 2^1 = 2
Rollout 71, Act: 2.127
Let me check.\n\nSuppose c is odd. So we had before, after substitution, the equation
Rollout 68, Act: 2.107
k negative, x negative: thus kx positive). But our domain was indeed x in (-2,
Rollout 35, Act: 2.106
0 +19 +3=22. Which aligns with n +3.\n\nSimilarly, at
Rollout 19, Act: 2.098
+ 20)/2, then modulo k, we can compute k^2 modk=0
Top 16 Negative Activations
Rollout 27, Act: -1.613
Solving for the one-dimensional rms velocity:\n \[\n v_{z, \text{rms
Rollout 83, Act: -1.617
sum of every three consecutive vertices is a multiple of 3, we start by analyzing the residues modulo
Rollout 83, Act: -1.617
of a regular nonagon with the digits 1 through 9 such that the sum of every three consecutive
Rollout 2, Act: -1.623
, then combine the results using the Chinese Remainder Theorem.\n\n1. **Modulo 8 Calculation
Rollout 17, Act: -1.623
pi} \left( \frac{3(\mathbf{\mu}_p \cdot \mathbf
Rollout 92, Act: -1.626
To find the refractive power of the lenses, we use the lens formula \( \frac{1}{
Rollout 40, Act: -1.632
consists of a counterclockwise rotation by \(\pi/4\) radians about the origin followed by
Rollout 80, Act: -1.678
, \text{cm}\n \]\n \[\n \text{Volume} = \frac
Rollout 29, Act: -1.678
at rest, the pressure at a depth \( h \) is given by \( P = \rho gh
Rollout 20, Act: -1.694
\) must be the least common multiple (LCM) of 43, 42,
Rollout 20, Act: -1.703
, \( n \) must be the least common multiple (LCM) of 43,
Rollout 16, Act: -1.719
omial theorem and swapping the order of summation, we get:\n\n\[\nE[T] = \sum
Rollout 92, Act: -1.729
0.615}\n \]\n \[\n \frac{1}{f} =
Rollout 20, Act: -1.734
) must be the least common multiple (LCM) of 43, 42,
Rollout 26, Act: -1.803
4.8 \, \text{K} = 296.0 \, \text
Rollout 29, Act: -1.882
= \rho gh \), where \( \rho \) is the fluid density, \( g \)

DOWN_PROJ

Top 16 Positive Activations
Rollout 20, Act: 3.054
43 and 47, then take LCM. Let me first work modulo 43.\n\n
Rollout 20, Act: 2.678
into 2 * 23.\n\nSo LCM(42,46) = 2
Rollout 20, Act: 2.662
works both mod43 and47 is LCM(1806,2162
Rollout 74, Act: 2.555
. Let me define the resultant field as the vector sum of H_e and the ship's field.\n\nLet
Rollout 20, Act: 2.498
62. So, n must be the least common multiple of 1806 and 2
Rollout 20, Act: 2.488
,46 mod47). Therefore, LCM(47,46)=216
Rollout 20, Act: 2.456
.\n\nTherefore, the minimal such n is the least common multiple of the minimal n that works mod 4
Rollout 20, Act: 2.455
2 and43.\n\nSo,n should be LCM(42,43)=180
Rollout 20, Act: 2.452
Thus, \( n \) must be the least common multiple (LCM) of 43,
Rollout 20, Act: 2.428
p, n must be a multiple of the least common multiple of the multiplicative orders of all p modulo
Rollout 20, Act: 2.426
47 and mod46. n=LCM(46,47)=216
Rollout 20, Act: 2.411
2162.\n\nTotal minimal n is LCM(1806,2162
Rollout 20, Act: 2.400
. Therefore, the minimal such n is the least common multiple of 42 and 46.
Rollout 20, Act: 2.379
conditions, n must be ��0 mod LCM(43,42)=180
Rollout 91, Act: 2.365
= (2, 0)\n\nSo the angle between BA and BC is angle at B.\n\nSo let
Rollout 20, Act: 2.348
- 1. So, meaning that the least common multiple of all possible orders (i.e., the
Top 16 Negative Activations
Rollout 20, Act: -2.367
, p^{kn} �� 1 mod 47 for all primes p ≠ 47
Rollout 66, Act: -2.369
parameterization, sometimes you have m and n coprime but lead to a scaled triplet. Wait,
Rollout 66, Act: -2.374
n2. Since the legs need to be coprime, which occurs when m and n are cop
Rollout 66, Act: -2.384
c = p.\n\nSince the sides are pairwise coprime, it's a primitive Pythagorean triple
Rollout 64, Act: -2.385
)/2 \), then for \( A \) and \( B \) to be integers (to ensure
Rollout 71, Act: -2.392
Similar for k2. Since 3c2 and k2 sum to 8044 (
Rollout 66, Act: -2.429
rime with both. Hence the legs are coprime only if they are primitive. But if the
Rollout 21, Act: -2.431
1. But if f(x) must map to Z, then if x is increased by 1,
Rollout 71, Act: -2.486
c2 +8044 is a perfect square.\n\nGiven that 8044 -3
Rollout 66, Act: -2.490
of primitive triples, m and n must be coprime. Therefore, to have gcd(m, n
Rollout 66, Act: -2.582
rime with each leg if the legs are coprime. So since hypotenuse c = m
Rollout 83, Act: -2.610
i+3} �� 0 (mod 3)\n\nSubtracting the first equation from the
Rollout 89, Act: -2.677
Therefore, the total number of children must be a multiple of 5, as 1/5th
Rollout 71, Act: -2.688
44 -3c2 to be a perfect square.\n\nAlternatively, since 8044 itself
Rollout 66, Act: -2.874
. Wait, no, all three are pairwise coprime? Wait, no. Wait, hypoten
Rollout 66, Act: -2.958
primitive Pythagorean triple, two legs are coprime, and hypotenuse is not copr

Layer 28

GATE_PROJ

Top 16 Positive Activations
Rollout 30, Act: 3.578
problem asks to "evaluate the infinite sum", likely as a fraction. But let's confirm.\n\nTherefore,
Rollout 3, Act: 3.459
olar mass and R? No, the question is to compute the speed, so they must assume a temperature
Rollout 33, Act: 3.452
than k in the first k positions, which is equivalent to saying that the firstk positions cannot be a
Rollout 96, Act: 3.420
translation and rotation.\n\nWait, because the coordinate systems are rotated and shifted. The original coordinate system is transformed
Rollout 25, Act: 3.411
make contact with the Earth?"\n\nSo, the question is: given the same approach speed at infinity and the
Rollout 22, Act: 3.381
333 �� ?\n\nWe can write that as 18,168.77
Rollout 71, Act: 3.352
-2011, which can be written as ab + c(a + b) = -2
Rollout 26, Act: 3.329
.\n\nBut how is the question phrased? "the calculated standard molar enthalpy... at
Rollout 99, Act: 3.327
What's the decimal value of x3?" Which is confusing, since two iterations would normally get to x
Rollout 33, Act: 3.310
number greater than k in the first k positions, which is equivalent to saying that the firstk positions cannot
Rollout 10, Act: 3.308
exist. As a contest problem, the solution may be found in realizing the periodicity failed, but that
Rollout 25, Act: 3.303
contact with the Earth?"\n\nSo, the question is: given the same approach speed at infinity and the same
Rollout 88, Act: 3.294
the entire activation output of this first layer", i.e., if you flatten all filters and channels into a
Rollout 14, Act: 3.271
side is $I^3$, perhaps writing that as $I \cdot I \cdot I$, and
Rollout 0, Act: 3.261
20!.\n\nHence, the question is equivalent to: how many pairs (a, b)
Rollout 94, Act: 3.254
) is not entangled.\n\nSo the question is "which set of states mentioned below are only entangled
Top 16 Negative Activations
Rollout 70, Act: -1.342
0}^k d_i * 10^i = 2 * sum_{i=0
Rollout 70, Act: -1.345
0}^k d_i \cdot 7^i \).\n - The base-10
Rollout 70, Act: -1.364
=0}^k d_i (10^i - 2*7^i) =
Rollout 70, Act: -1.382
a base-10 number, equal twice \( N \). To find the largest such \( N \
Rollout 96, Act: -1.382
87/28561) +5s= -4106/16
Rollout 70, Act: -1.401
=2N.\n\nThus,\n\nd3*1000 + d2*100 +
Rollout 70, Act: -1.406
0}^k d_i \cdot 10^i \).\n - We need \( D
Rollout 70, Act: -1.439
2N.\n\nThus,\n\nd3*1000 + d2*100 +d
Rollout 70, Act: -1.473
=0}^k d_i (10^i - 2 \cdot 7^i
Rollout 70, Act: -1.533
0}^k d_i * 10^i. So, we need the same digits d
Rollout 70, Act: -1.549
_0. Then D=d_k*10^k +...+d_1*10
Rollout 70, Act: -1.596
d0.\n\nD= d3*10^3 +d2*10^2 +
Rollout 70, Act: -1.626
2N.\n\nTherefore,\n\nd_k*10^k + ... + d_1*10
Rollout 70, Act: -1.627
}^k d_i \cdot 10^i = 2 \sum_{i=0
Rollout 70, Act: -1.634
_0 = 2*(d_k*7^k + ... + d_1*7 +
Rollout 70, Act: -1.854
}^k d_i \cdot 10^i \).\n - We need \( D =

UP_PROJ

Top 16 Positive Activations
Rollout 36, Act: 3.159
Let's compute 18,8642:\n\nFirst compute 18,000
Rollout 4, Act: 3.084
000, since 5025*1000=5,025
Rollout 36, Act: 3.076
58.12 because 1582=24,964 and 1
Rollout 36, Act: 3.036
2):\n\nSince 19,0002 = 361,000,
Rollout 36, Act: 2.897
0\n\nCompute 19,3602=?\n\n19,360^2
Rollout 71, Act: 2.878
d <= 25 (since 252=625 and 262=
Rollout 36, Act: 2.824
2=?\n\n19,360^2= (19,000 +
Rollout 36, Act: 2.789
to 75.0 because 752=5,625, very close.\n\n
Rollout 22, Act: 2.773
compute this.\n\nSince 1,7002 = 2,890,00
Rollout 71, Act: 2.754
be up to 25 since 252=625 <=670.3
Rollout 11, Act: 2.731
e.g., k = 2i (modulus 2), then c =1/( (2
Rollout 38, Act: 2.725
27 because 527*527=?\n\nWell, 5002=
Rollout 71, Act: 2.716
=> c>=45 (since 442=1936, 452
Rollout 38, Act: 2.700
055. Compute 10542=(1055-1)^2=
Rollout 71, Act: 2.692
4.81, no. 442=1936, 452
Rollout 22, Act: 2.683
0). \n\nSince 1,6942 = 2,869,63
Top 16 Negative Activations
Rollout 32, Act: -1.487
pmatrix} 0 & 1 \\ -\omega^2 & -\omega \end{
Rollout 28, Act: -1.488
B h + 2C k - C b = 0\n \end{cases}\n
Rollout 51, Act: -1.492
integral becomes, after variable shifting y =x -θ, integral becomes:\n\n∫02π sin8
Rollout 43, Act: -1.499
|0\rangle\)\n - \(-\gamma|1\rangle \otimes |0
Rollout 28, Act: -1.515
=0\n\nFactor terms in (2k -b) and (2h -a):\n\n(2
Rollout 28, Act: -1.533
4a2y2 -4a2b y =4a2y(y -b )\n\n
Rollout 28, Act: -1.535
b = 0 ⇒ C(2k - b) + B h = 0\n\nSo we
Rollout 28, Act: -1.540
0 ... (1)\n\nC(2k - b) + B h = 0 ... (2
Rollout 64, Act: -1.544
terms are still integers. Therefore, the only constraints are that a1 and a2 are positive integers.\n\n
Rollout 12, Act: -1.559
= [ t / sqrt( (t2 - ν2)(1 - t^2) ) ]
Rollout 75, Act: -1.581
should be possible. Wait, the only constraint here is that S �� 3 mod4. Let
Rollout 28, Act: -1.592
3(u -2)^2 -12 + v2 +9=0 → 3(u -
Rollout 12, Act: -1.607
[ t / sqrt( (t^2 - ν^2)(1 - t^2) )
Rollout 28, Act: -1.610
2 + 2a2 b k -4 b2 h2 + 2a b2 h
Rollout 28, Act: -1.617
(-2A a^2 - 2C b^2) + B ab = 0 \
Rollout 45, Act: -1.767
4 = f(-1) + 5 - b => f(-1) = -4 -5

DOWN_PROJ

Top 16 Positive Activations
Rollout 60, Act: 2.870
6 Hz.\n\nBut why the initial approximate calculation gave something like 5.92 Hz, but
Rollout 23, Act: 2.854
5 Bq, which is option (D) 117.555, close to
Rollout 70, Act: 2.670
153.\n\nBut N=35m +13. Thus,\n\n105d1
Rollout 2, Act: 2.658
109 in both cases. So The answer is109. Huh.\n\nBut wait,
Rollout 36, Act: 2.623
0\n\nCompare to our Δ=375,662,592. It's
Rollout 36, Act: 2.582
00\n\nCompare to our Δ=375,662,592. It
Rollout 39, Act: 2.567
4. Hmmm. 27 isn't 4, so seems off. So I must be
Rollout 91, Act: 2.539
C|=2 regardless of t?\n\nSo the answer is 2?\n\nBut wait, wait. So according
Rollout 87, Act: 2.513
62. That's not matching 3.795.\n\nAlternatively, different combination.\n\nTest another
Rollout 30, Act: 2.474
5 *2014 )\n\nSo the answer is 1 + 1/(2015
Rollout 58, Act: 2.449
7171. Thus, the answer should be \boxed{2037171
Rollout 11, Act: 2.417
] + [ r2 + 2/r cosθ ]= 2r cosθ +1/r
Rollout 70, Act: 2.410
102.\n\nBut N=35m +32.\n\nThus,\n\n105d1
Rollout 8, Act: 2.410
.246.\n\nThus, the answer would be 31 +126=15
Rollout 90, Act: 2.402
the deduced answer, yes. So the answer is \boxed{1008919
Rollout 38, Act: 2.384
24=61.\n\nTherefore, the answer is \boxed{61}。\n\n**Final Answer
Top 16 Negative Activations
Rollout 89, Act: -2.458
Wait, wait a minute. Let me read the problem again. "Exactly 2/5th of
Rollout 81, Act: -2.461
/121 -2601/121= (6171 -2
Rollout 96, Act: -2.464
*37=481 →Total 109,200 +481
Rollout 9, Act: -2.468
(10 -5)/(10 - (10 - h)) = 5/h.\n\nTherefore
Rollout 81, Act: -2.471
, possibly error in methods complicated.\n\nLet me step back. If both approaches correctly model tan(theta)=2
Rollout 99, Act: -2.474
(-4)/(-3.456315) �� 1.157
Rollout 99, Act: -2.496
in the sign for the denominator. Let me go back step by step.\n\nSo, step 1:
Rollout 79, Act: -2.503
No.\n\nThis is really frustrating. Wait, going back to anagram of "minor changes": M,
Rollout 59, Act: -2.517
/28,\n\nP(X=3) = 1/7 = 4/28.\n\n
Rollout 87, Act: -2.539
is integrating 1/sqrt(ξ2 + ��2 + ��2}) over the unit
Rollout 56, Act: -2.575
44\n\nThus sqrt[196/144 + 363/14
Rollout 1, Act: -2.594
2) d �� 1.414d. The mutual distance between centers is d.
Rollout 22, Act: -2.632
number (6.02214076e23 molecules/mol):\n\n0.0
Rollout 70, Act: -2.673
47=6. Then 6 divided by 7=0, so digits should be 3
Rollout 61, Act: -2.673
0} \) whose binary expansions have exactly two 1's, we need to find the probability that
Rollout 97, Act: -2.715
2 moles of N2O4 and 1 mole of O2 are formed.\n\nSo,

Layer 29

GATE_PROJ

Top 16 Positive Activations
Rollout 22, Act: 4.097
= that’s negative, maybe.\n\nWait, better way: 145,458.
Rollout 20, Act: 4.040
Because4^7=,wait no, better way. Wait,2^14≡4 mod
Rollout 73, Act: 3.997
n+1}\n\nWait, perhaps a different way:\n\nLooking term by term:\n\nFirst term of 2
Rollout 41, Act: 3.995
first digit). Wait, no.\n\nWait, perhaps better to split into two cases: when one of the
Rollout 20, Act: 3.957
. Because4^7=,wait no, better way. Wait,2^14≡4
Rollout 13, Act: 3.917
approach. Wait, maybe I should try direct multiplication:\n\n1.2544\n*1.
Rollout 70, Act: 3.888
^{-1}*2 mod5. Wait, perhaps better to use previous equations.\n\nSo, if d0
Rollout 72, Act: 3.862
2. No, better to use exact formula. cos(72°) = (sqrt(
Rollout 22, Act: 3.847
that’s negative, maybe.\n\nWait, better way: 145,458.9
Rollout 25, Act: 3.845
me square both sides first. Wait, perhaps a better substitution.\n\nLet’s let x = r2,
Rollout 96, Act: 3.838
2x +5y=90\n\nSolving for variables:\n\nAlternatively, solve for one variable from
Rollout 8, Act: 3.834
BB... Wait, GGRR would actually be similar: P(5,2)*P(5,
Rollout 40, Act: 3.829
�^n,\n\nWait, let me group the terms:\n\nFirst term: 5 ��^n\n\nSecond
Rollout 9, Act: 3.795
dx\n\nWait, wait subtract the lower from upper:\n\n= ��[0 to10] [
Rollout 36, Act: 3.769
Wait, wait, hold on:\n\nWait, more accurately, (s2 + 1886
Rollout 16, Act: 3.760
sapling is not yet converted.\n\nWait, perhaps another approach. Since each day, each remaining sapling
Top 16 Negative Activations
Rollout 77, Act: -1.709
,52)$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this problem here where I
Rollout 12, Act: -1.712
P. Turan]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to prove that the function
Rollout 47, Act: -1.713
of $N$?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so there's this problem about a train
Rollout 89, Act: -1.722
are in the class?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me see here. So the problem
Rollout 1, Act: -1.723
cal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that there exists
Rollout 74, Act: -1.731
and $\theta$.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to derive an expression for
Rollout 94, Act: -1.743
(D) a,b\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to determine which of these
Rollout 97, Act: -1.745
mathrm{Y}$?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I have this decomposition
Rollout 80, Act: -1.764
of $N$ ?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure out this problem
Rollout 29, Act: -1.764
rho gH$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure out this physics
Rollout 42, Act: -1.769
-5-ene\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to figure out the product
Rollout 44, Act: -1.774
{T}_{2}$\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to figure out which of
Rollout 27, Act: -1.783
the line of sight).\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I've got this problem here where
Rollout 23, Act: -1.798
55 Bq\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this problem step by step
Rollout 57, Act: -1.806
. Csaszar*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that if we
Rollout 98, Act: -1.806
(D) 2\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to figure out how many

UP_PROJ

Top 16 Positive Activations
Rollout 37, Act: 2.922
5 and their multiplicities. But perhaps for each prime p ≤ 2015, the
Rollout 37, Act: 2.799
seems complicated if we need to check it for each prime.\n\nAlternatively, maybe the determinant can be expressed as
Rollout 33, Act: 2.794
first k terms include at least one number greater than k. Let me try to unpack this problem and figure
Rollout 93, Act: 2.775
, but the power sum symmetric polynomials of the roots. If x_k = cos(πk
Rollout 90, Act: 2.768
's a corresponding N �� -897M mod 1009.\n\nBut N is
Rollout 37, Act: 2.691
≤ 2015, the exponent of p in det(A) is at least the exponent of
Rollout 37, Act: 2.686
for any prime p ≤ 2015, the exponent of p in det(A) is at
Rollout 66, Act: 2.677
m2 - n2, b = 2mn, c = m2 + n2. Let
Rollout 0, Act: 2.651
b\). So for each such coprime pair where \(a < b\), we have a
Rollout 33, Act: 2.644
should be at least one number that's bigger than k. And this condition needs to hold for all k
Rollout 20, Act: 2.640
43.\n\nSo, we need for each prime p, and each k ≥ 1, σ
Rollout 20, Act: 2.610
isors of a. So, if a is a prime p, σ(p) = 1 + p
Rollout 44, Act: 2.604
it's isothermal, the work would be nRT ln(V1 / V2). Or, if
Rollout 37, Act: 2.600
orem, it suffices to show that for any prime p ≤ 2015, the exponent
Rollout 33, Act: 2.588
numbers 1 through 6 such that for each k from 1 to 5, the first k
Rollout 93, Act: 2.585
Wait, another thought: when dealing with sums over roots of unity or trigonometric sums, sometimes symmetry
Top 16 Negative Activations
Rollout 53, Act: -2.847
be answer.\n\nHence, the area is \boxed{\dfrac{27}{2}}.\n\n
Rollout 66, Act: -2.848
=182. Thus, answer is \boxed{182}.\n\nTherefore, despite m
Rollout 13, Act: -2.860
7.12\n\]\n\n\[\n\boxed{187.12}\n\]
Rollout 8, Act: -2.861
= 157\).\n\n\[\n\boxed{157}\n\]<|im_end|>
Rollout 43, Act: -2.862
be \boxed{C}.\n\n\n**Final Answer**\n\boxed{C}\n\n<|im_start|>answer\nThe problem
Rollout 50, Act: -2.863
112\).\n\n**Final Answer**\n\boxed{112}\n\n<|im_start|>answer\nGiven the
Rollout 5, Act: -2.872
5 is 181.\n\n\[\n\boxed{181}\n\]<|im_end|>
Rollout 12, Act: -2.895
\nu < 1 \).\n\n\[\n\boxed{f(\nu) \text{ is mon
Rollout 68, Act: -2.898
= 501 \).\n\n\[\n\boxed{501}\n\]<|im_end|>
Rollout 32, Act: -2.921
is definitely 4018.\n\nTherefore \boxed{4018}.\n\n**Final Answer
Rollout 76, Act: -2.927
= 929\).\n\n\[\n\boxed{929}\n\]<|im_end|>
Rollout 78, Act: -2.956
n = 49 \).\n\n\[\n\boxed{49}\n\]<|im_end|>
Rollout 16, Act: -2.991
9953\n\]\n\n\[\n\boxed{789953}\n\]
Rollout 50, Act: -3.045
112\n\]\n\n\[\n\boxed{112}\n\]<|im_end|>
Rollout 61, Act: -3.057
80 = 913 \).\n\n\boxed{913}<|im_end|>
Rollout 59, Act: -3.113
= 515\).\n\n\[\n\boxed{515}\n\]<|im_end|>

DOWN_PROJ

Top 16 Positive Activations
Rollout 14, Act: 2.900
the 1/x integral between \ the tails.\n\nAlternatively, use Hölder inequality in the form:\n\n(
Rollout 87, Act: 2.815
by something. Not sure, but some combination.\n\nAlternatively, recall that potential at center is different—the integral
Rollout 79, Act: 2.803
anagram of "monarchesign"? No. Alternatively, "morchaseingn" – which
Rollout 46, Act: 2.796
, while we are asked for second time derivatives.\n\nAlternatively, applying U in context of Lagrangian formal
Rollout 49, Act: 2.792
If A is well-ordered by some relation, but need to get a sequence, but I can't
Rollout 37, Act: 2.764
if a1 and a2 are consecutive integers, but we have no info.\n\nWait, let's plug
Rollout 79, Act: 2.747
't possible. \n\nNext, "Chancers" needs two C's. No. \n\nNext, "
Rollout 49, Act: 2.738
ity with hereditary Lindelöfness?\n\nAlternatively, consider as X is hereditarily Lindel
Rollout 87, Act: 2.737
multiple integrations substituting hyperbolic substitutions.\n\nAlternatively, in Feynman's lectures, they solve
Rollout 37, Act: 2.734
of finite differences but I don't remember well.\n\nAlternatively, seeing that for the. statement to be true
Rollout 87, Act: 2.723
. The cube has edges, corners, etc.\n\nAlternatively, integrate in spherical coordinates but adjusting for the cube
Rollout 46, Act: 2.713
�2 x / something... Wait, no.\n\nAlternatively, perhaps confusion between Laplace's and gradient.
Rollout 93, Act: 2.710
directly because the relations depend on parity of k.\n\nAlternatively, for all x_k we have T_{1
Rollout 87, Act: 2.698
result in terms with inverse trigonometric functions.\n\nAlternatively, since each step involves multiple integrations substituting
Rollout 79, Act: 2.687
, "Mechanic's Sign" – no.\n\nAlternatively, maybe "Minor changes" becomes "Change minor
Rollout 79, Act: 2.686
".\n\nAlternatively, think of "Mosaic" – but not enough letters.\n\nAlternatively, maybe the 4
Top 16 Negative Activations
Rollout 41, Act: -2.156
Numbers where both digits are non-zero. Then, we can arrange those two digits freely, but ensuring no
Rollout 41, Act: -2.171
the total number according to the alternative approach (if leading zeros are allowed) is 640-
Rollout 41, Act: -2.179
3 numbers. Therefore, total numbers for this case would be 9 * 3 = 2
Rollout 77, Act: -2.190
6)=12, which is LCM(4,6)=12. Indeed. Then f
Rollout 32, Act: -2.190
�2 - ��)=4 (same as above since 1 - ��2 - ��
Rollout 41, Act: -2.195
two different digits is 243, plus the numbers with all three digits the same, which is
Rollout 94, Act: -2.215
able.\n\nHence, again, my previous conclusion holds.\n\nTherefore, the correct answer is option (B
Rollout 39, Act: -2.215
10. But in this smaller case, how would it break down with subtraction? Well, taking us
Rollout 3, Act: -2.219
Wait, but I might have to verify which one is the actual mean speed. Let me think.\n\nMean
Rollout 60, Act: -2.239
5.92 Hz, but using exact fractions gives exactly 6 Hz? Probably because I approximated
Rollout 48, Act: -2.245
16=7 with carry 1. So we write 7.\n\nNext (tens digit of
Rollout 95, Act: -2.247
. There is a discrepancy here. So which is correct?\n\nAlternatively, there's a confoundment.
Rollout 63, Act: -2.293
1}{2}f(2,2) \), substitute into above:\n\n\[\nf(2
Rollout 32, Act: -2.363
gives 669*6. Then the remaining two terms (M and M2) would give
Rollout 41, Act: -2.401
, plus the numbers with all three digits the same, which is 9. So in total,
Rollout 97, Act: -2.721
5 e^{-kt}\n\nWhich is the same result as when I did it via ODE and stoich

Layer 30

GATE_PROJ

Top 16 Positive Activations
Rollout 41, Act: 2.478
,000 that have at most two different digits is \boxed{927}.\n\n<|im_start|>
Rollout 41, Act: 2.246
- **Case 2**: Exactly two different digits.\n - **Subcase A**: Neither digit
Rollout 41, Act: 2.246
0,000 have at most two different digits, we consider numbers of different lengths (1-digit
Rollout 41, Act: 2.220
0,000 have at most two different digits?\n\nSo if I just think of all numbers from
Rollout 41, Act: 2.168
to count those numbers that use at most two different digits. For example, 111 has one
Rollout 41, Act: 2.148
10 numbers.\n\nFor numbers with exactly two different digits: Choose two digits (C(10,
Rollout 41, Act: 2.084
same (9 numbers) plus numbers with exactly two different digits (243) as computed above.
Rollout 41, Act: 2.077
each of these numbers must have at most two different digits. So, numbers like 111 would
Rollout 41, Act: 2.069
0,000 have at most two different digits?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so
Rollout 41, Act: 2.068
- **Case 2**: Exactly two different digits.\n - **Subcase A**: Neither digit
Rollout 41, Act: 1.977
How many four-digit numbers have at most two different digits?\n\nAgain, we split into cases:\n\nCase
Rollout 41, Act: 1.968
10 numbers.\n\nFor numbers with exactly two different digits: Choose two digits (C(10
Rollout 41, Act: 1.923
numbers (including leading zeros) where only two distinct digits are present, then subtract 1 (the number
Rollout 41, Act: 1.917
0,000 have at most two different digits? Hmm, okay. Let me try to break
Rollout 41, Act: 1.902
leading zeros):\n\nThe total numbers with at most two different digits would be:\n\nNumber of monochromatic numbers
Rollout 41, Act: 1.898
need to count those numbers that use at most two different digits. For example, 111 has
Top 16 Negative Activations
Rollout 44, Act: -3.950
), then ΔV = V1 - V2 is negative. So work done on the gas is -
Rollout 22, Act: -3.956
, the mean molecular speed, right. The formula I remember is something like the square root of (3
Rollout 69, Act: -3.974
^i)]\n\nI need to evaluate that infinite sum and express it as 1/(a + ��
Rollout 81, Act: -3.989
/17 (negative slope). Then, difference is (-h/17 - h/3).\n\n
Rollout 99, Act: -3.990
3. Hmm, okay. \n\nFirst, I remember that the Müller method is a numerical method used to
Rollout 74, Act: -4.005
's magnetic field differently, causing the compass deviation.\n\nTo model this, I should consider the Earth's magnetic
Rollout 27, Act: -4.012
start breaking this down.\n\nFirst, I need to recall what rms speed means. Root mean square speed,
Rollout 12, Act: -4.020
out how to approach this. \n\nFirst, I remember that a function is monotonically decreasing if its
Rollout 28, Act: -4.033
break this down step by step.\n\nFirst, I remember that a conic is defined by five points,
Rollout 48, Act: -4.037
AEE0. Alright, first, I should recall the structure of an IPv4 header to parse each
Rollout 54, Act: -4.059
ethylcyclohexanol's pKa—probably similar to cyclohexanol, maybe a bit
Rollout 36, Act: -4.067
456 - 55584):\n\n147456 - 55
Rollout 17, Act: -4.087
**I**|. \n\nFirst, I need to recall the formula for the magnetic field of a dipole
Rollout 3, Act: -4.126
of radon gas. Hmm, first, I remember there's a formula for the root mean square speed
Rollout 92, Act: -4.176
01626. So adding these two: 5.405405 -
Rollout 29, Act: -4.243
heights h and H.\n\nFirst, I need to recall what gauge pressure is. Gauge pressure is the pressure

UP_PROJ

Top 16 Positive Activations
Rollout 55, Act: 3.621
, (2,1) entry: 0* stuff would be 0. Similarly, (2
Rollout 53, Act: 3.598
/2,0 )\n\nFirst term: x1 y2 = (-sqrt73 /2)(0
Rollout 86, Act: 3.527
0 )=0.\n\nAt0: 0*( -1 + (1/3)*1 )
Rollout 73, Act: 3.507
4? Because when i=0, 0*2^0=0. So sum_{i
Rollout 40, Act: 3.460
1 )].\n\nThen plugged into term [10/(1 - �� )]*(1 +i).
Rollout 86, Act: 3.447
/3) =0.\n\nAt0:0*(-1/3)=0.\n\nThus, uv=
Rollout 33, Act: 3.431
, but if you already have the first two elements as 1 and 2 (in some order),
Rollout 53, Act: 3.401
2)(0 ) = 0\n\nx2 y3 = ( 0 )*(18/s
Rollout 52, Act: 3.396
Maybe check the math.\n\nExpress the sum as average * number of terms. The average of the sequence from
Rollout 56, Act: 3.388
implify term by term:\n\nFirst term: 0*( -√3/2 - ��3/
Rollout 73, Act: 3.383
by term.\n\ni=0: 10*1=10\n\ni=1:9
Rollout 18, Act: 3.362
take (k, n) = a*(0, x) + k*(1, y) =
Rollout 86, Act: 3.355
�x cosx dx.\n\nEvaluate first term: - ( (pi/2)^2 * cos(pi
Rollout 56, Act: 3.354
168 )\n\nArea:\n\n= | x3(y4 - y5) + x4(y5
Rollout 53, Act: 3.353
*0 =0\n\nPoint3: x3 y2: (23/(2 sqrt73
Rollout 86, Act: 3.345
dx.\n\nEvaluate terms:\n\nAt pi/2: - (1/2)(pi/2) cos
Top 16 Negative Activations
Rollout 39, Act: -2.362
21 terms in total? Wait, no. Wait, let me check. If the first term is
Rollout 22, Act: -2.379
e-23 kg? Wait, no. Wait, 0.002016
Rollout 20, Act: -2.386
the exponent of the multiplicative group modulo m. But the multiplicative group modulo a prime m is cyclic
Rollout 11, Act: -2.393
/b + a/c is the reciprocal of q? Wait, if we take reciprocal of each term in q
Rollout 40, Act: -2.400
/(1 - ��)*(1 + i). Wait, earlier:\n\nz_{150} =
Rollout 83, Act: -2.401
1296. But wait, no. Wait: if each residue is assigned to exactly one
Rollout 42, Act: -2.406
adjacent carbons are not on the rings themselves. Wait, no. Let me think again. The alcohol
Rollout 20, Act: -2.421
(m - 1)? Wait, no, wait: The exponent of a cyclic group is equal to the
Rollout 26, Act: -2.422
.71 kJ/mol\n\nWait a second. But hold on. The enthalpy of formation
Rollout 80, Act: -2.440
0^26 grams? Wait, hold on: 12.8 * 7.2
Rollout 86, Act: -2.444
should integrate to the same value because of symmetry. Wait, let me verify.\n\nBut actually, let me
Rollout 46, Act: -2.456
, that gives the acceleration as the gradient. But wait, in rotating frames, there are additional terms,
Rollout 26, Act: -2.471
71 kJ/mol\n\nWait a second. But hold on. The enthalpy of formation of
Rollout 40, Act: -2.476
Wait, actually factoring out ��^n,\n\nWait, let me group the terms:\n\nFirst term:
Rollout 69, Act: -2.536
0/9°, etc.? Wait, wait: wait, when i=0: 30
Rollout 9, Act: -2.554
, EM is the top side of the square. Wait, hold on. \n\nWait, in the square

DOWN_PROJ

Top 16 Positive Activations
Rollout 14, Act: 3.386
) sqrt(A ) +K/A.\n\nSet derivative with respect to A to zero.\n\ndh/dA=
Rollout 71, Act: 3.268
to sums and products of its roots. Let me recall them for a cubic equation. If the roots are
Rollout 52, Act: 3.215
formula for the sum of consecutive numbers. Let me remember. If you have a sequence starting at a and
Rollout 14, Act: 3.205
fixed $J$ and $K$. If I can show that the maximum possible value of $I^
Rollout 38, Act: 3.171
sqrt=527 because 527*527=?\n\nWell, 50
Rollout 51, Act: 3.123
π. Hmm, okay. Let me see how to approach this.\n\nFirst off, the integral involves an
Rollout 99, Act: 3.113
10^0.334 �� 2.157), so 0
Rollout 19, Act: 2.999
2} \right)^2 \). Let me recall that formula. Yes, that’s right: \(
Rollout 40, Act: 2.976
Or perhaps using complex numbers? Hmm. Let me recall that rotations can be represented using complex numbers by multiplying
Rollout 14, Act: 2.962
with exponents that are conjugate. Let me recall Hölder's inequality: for $p$ and
Rollout 21, Act: 2.913
like sin(π f(x)) = 0 would force f(x) to be integer, but the
Rollout 66, Act: 2.897
in two ways: (leg1 * leg2)/2, or (hypotenuse * altitude)/
Rollout 22, Act: 2.885
,333 *2,869≈18,157, 6,
Rollout 56, Act: 2.860
a right-angled triangle is (base * height)/2. Here, the legs are A0A
Rollout 87, Act: 2.857
this kernel—maybe tabulated somewhere. Let me recall. In fact, in some textbooks, like Griffith
Rollout 82, Act: 2.850
algorithm for a mixed-radix system? Let me recall how Huffman coding works. For a binary code,
Top 16 Negative Activations
Rollout 22, Act: -2.519
which one they want.\n\nWait, the question does say "mean molecular speed," so that would be the
Rollout 31, Act: -2.529
-1) =0.\n\nSo the cubic factors as (x +1)(x^2 -x
Rollout 75, Act: -2.532
and then combining them, but no, the problem states that the sum of the three squares can be expressed
Rollout 49, Act: -2.539
subspace is σ-compact. So even subspaces that are not closed or not nice? For example
Rollout 76, Act: -2.541
-3, cols1-2.\n\nCombined, rows 1-2: cols 1-3
Rollout 59, Act: -2.548
(3/28)^2?\n\nWait, but wait, so originally each of their trials is independent
Rollout 8, Act: -2.549
adjacent). But the way the problem is phrased: "all the red cards laid out are adjacent
Rollout 79, Act: -2.560
, N). So perhaps "in" is just part of the clue or used differently.\n\nAlternatively, maybe
Rollout 84, Act: -2.569
twice the original purchased amount? Hmm, the problem says "twice as much". When we say someone
Rollout 79, Act: -2.586
etc. "Noticed" might be the anagram indicator. "Minor changes" could be the letters
Rollout 46, Act: -2.586
Alternatively, if the Coriolis term is part of the potential, but it's not. The Cor
Rollout 43, Act: -2.590
to |1⟩. So in the input state, we have:\n\nα|00⟩ becomes α
Rollout 42, Act: -2.601
cyclopropane-derived) and a four-membered (cyclobutane-derived) ring
Rollout 54, Act: -2.765
. 4-methylcyclohexan-1-olate (bulky alkoxide, despite
Rollout 79, Act: -2.913
N). So perhaps "in" is just part of the clue or used differently.\n\nAlternatively, maybe only
Rollout 36, Act: -3.018
3s^4\n\nTotal s2 terms: 1,728 +37,

Layer 31

GATE_PROJ

Top 16 Positive Activations
Rollout 18, Act: 4.487
5). But let me double-check once again in case of mistakes.\n\nFirst, check that H_{
Rollout 94, Act: 4.478
verify in case I made a mistake:\n\nWait, in state (d): coefficients are a=1,
Rollout 72, Act: 4.385
even positive integers. Wait, but let's check in the problem statement. Are we supposed to answer that
Rollout 94, Act: 4.303
is correct.\n\nWait, let me cross-verify in case I made a mistake:\n\nWait, in state
Rollout 45, Act: 4.213
x + y = 0, but let's see.\n\nAlternatively, think about setting x or y to
Rollout 81, Act: 4.211
22/7\n\nWait a second, note in this approach we get 20h/(h
Rollout 39, Act: 4.176
1 and then a 0. Wait, the number of ones corresponds to the exponent minus 1
Rollout 99, Act: 4.175
to see if there are errors.\n\nWait, perhaps in the sign for the denominator. Let me go back
Rollout 45, Act: 4.174
4x2 -4.\n\nBut wait, but in the functional equation, axy(x + y)+
Rollout 96, Act: 4.168
90\n\nSolving for variables:\n\nAlternatively, solve for one variable from one equation and substitute.\n\nFrom
Rollout 42, Act: 4.152
bit complicated. Let's look at the answer choices and see if that can help.\n\nThe answer options are
Rollout 73, Act: 4.110
complicated. Let me try calculating for small n.\n\nWhen n=1: sum is 1*2
Rollout 94, Act: 4.109
me cross-verify in case I made a mistake:\n\nWait, in state (d): coefficients are a
Rollout 13, Act: 4.106
than 1.5625.\n\nBut the exact calculation gives around 1.573
Rollout 73, Act: 4.095
16398.\n\nBut wait, when I compute 16398 divided by
Rollout 45, Act: 4.086
b, c by plugging known values.\n\nLet me try plugging x = 1, y =
Top 16 Negative Activations
Rollout 83, Act: -1.570
8}\n\nGiven the constraint that the sum of every three consecutive digits must be divisible by 3, we
Rollout 49, Act: -1.587
\( X \) is \(\sigma\)-compact, then \( X \) is countable,
Rollout 10, Act: -1.594
understand the game first.\n\nPlayers take turns removing a square number of pebbles (1, 4
Rollout 49, Act: -1.594
ff space $X$ is $\sigma$ -compact, then $X$
Rollout 26, Act: -1.629
3\n \]\n Using R = 8.3 J/K/mol = 0.0
Rollout 49, Act: -1.639
space \( X \) is \(\sigma\)-compact, then \( X \) is countable
Rollout 10, Act: -1.642
analyze the game where players alternately remove a square number of pebbles. The key is to identify
Rollout 80, Act: -1.652
{Volume} = \frac{4}{3} \pi r^3 = \frac{4
Rollout 6, Act: -1.660
}$ with a constraint that there should exists a polynomial time algorithm that can **compute** the angel string
Rollout 10, Act: -1.701
the Game**:\n - Players take turns removing square numbers (1, 4, 9,
Rollout 10, Act: -1.756
the game where players alternately remove a square number of pebbles. The key is to identify losing
Rollout 10, Act: -1.783
bbles where $x$ is the square of any positive integer. The player who is unable
Rollout 57, Act: -1.796
1 \). Both sets are countable and thus have measure zero.\n2. **Closure in Density
Rollout 10, Act: -1.973
we analyze the game where players alternately remove a square number of pebbles. The key is to
Rollout 6, Act: -1.992
{P}_{angel}$ if there exists a polynomial $p : \mathbb{N}
Rollout 31, Act: -1.999
real numbers $x, y$. For any pair of real numbers $a_{0}, a_{1

UP_PROJ

Top 16 Positive Activations
Rollout 56, Act: 3.587
3√3/4 �� 1.299. Okay, so K must be at
Rollout 93, Act: 3.571
m=4:9/8=1.125. Not obvious. Maybe the closed-form
Rollout 14, Act: 3.504
the maximum possible ratio is approximately 2.25/8≈0.28. Wait,
Rollout 14, Act: 3.503
But this seems that the ratio is 2.25/8=0.2812
Rollout 14, Act: 3.447
but the maximum possible ratio is approximately 2.25/8≈0.28. Wait
Rollout 93, Act: 3.432
4:9/8=1.125. Not obvious. Maybe the closed-form is in
Rollout 14, Act: 3.426
this seems that the ratio is 2.25/8=0.28125
Rollout 14, Act: 3.410
)/(4/(k^3 ))=1/4. Which matches the ratio here. Hmmm.\n\n
Rollout 56, Act: 3.367
3√3 /4 ��1.299), as computed above. So the area
Rollout 93, Act: 3.285
/32≈1.03125, m=4:9/8=1
Rollout 44, Act: 3.225
2. The reversible work was 1.386. So yes, higher. But in
Rollout 38, Act: 3.204
.428x. So L≈1.857x < W≈3.4
Rollout 14, Act: 3.185
5/8=0.28125. Which indicates that equality is achieved in other cases
Rollout 81, Act: 3.140
11=46 and 4/11.\n\nBut the question states "What is the area
Rollout 56, Act: 3.122
3√3 /4 �� 1.299. Then we need to compute the areas
Rollout 93, Act: 3.104
1/8=9/8=1.125.\n\nHmm, so for m=4
Top 16 Negative Activations
Rollout 63, Act: -2.217
\) is zero. Otherwise, when both \( p \) and \( q \) are positive,
Rollout 38, Act: -2.226
.2 sqrt(x2 + W2)\n\nLet me express 6.2 as 31/
Rollout 9, Act: -2.228
25.\n\nPerfect.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(A
Rollout 93, Act: -2.234
014}}} since the clearest expression.\n\n**Final Answer**\n\boxed{\dfrac{1
Rollout 40, Act: -2.239
1 -omega), conjugate over denominator.\n\nWait, actually, we have 1/(1 -omega
Rollout 96, Act: -2.243
cosθ)\n\nBut I need to verify.\n\nWait, let's consider that. Let's say we have
Rollout 80, Act: -2.249
64 × 10^26 grams? Wait, hold on: 12.8
Rollout 52, Act: -2.251
013}.\n\n**Final Answer**\n\boxed{-2013}\n\n<|im_start|>answer\nTo
Rollout 80, Act: -2.334
km = 10^5 cm). Wait, 1 km = 1000 meters
Rollout 96, Act: -2.350
new x-axis with the original x-axis.\n\nWait, actually, when converting a point from original system to
Rollout 64, Act: -2.374
=z +1 + y\n\nWait, but I had previous substitution z=x -1. So x
Rollout 63, Act: -2.396
Otherwise, when both \( p \) and \( q \) are positive, the function satisfies the equation
Rollout 30, Act: -2.397
2, which is ≥2? Wait, hmm, the recursion for a_{n+1} is
Rollout 33, Act: -2.430
})! * (6 - km)!.\n\nWait, let me check with a concrete example. Take A
Rollout 91, Act: -2.450
c - b)/2, 0). Wait, but actually, if we have BC of length,
Rollout 56, Act: -2.537
4-2}= A4 A2. Wait, for n=4, n-2=2

DOWN_PROJ

Top 16 Positive Activations
Rollout 96, Act: 3.316
=132 +12y =>x=(132 +12y)/5\n\n
Rollout 7, Act: 3.306
32x=8N/3 => x=(8N)/(3*32)=N/
Rollout 96, Act: 3.090
=90 -12x\n\nTherefore y=(90 -12x)/5\n\nPl
Rollout 96, Act: 2.916
= -1134\n\nThus, y= -1134/169.\n\n
Rollout 95, Act: 2.869
(y/H)^2= k/N => y = H*sqrt(k/N).\n\n Hence, the height
Rollout 65, Act: 2.831
a2 =1995 c ⇒ c= a2 /1995.\n\nSimilarly,
Rollout 18, Act: 2.811
5b = -3 -4c => b = (-3 -4c)/5\n\nThis needs
Rollout 49, Act: 2.791
compact, we get a contradiction.\n\nWait, but maybe such a space X must be hereditarily Lind
Rollout 5, Act: 2.782
(x + 175)H [1 - 125/(75 + x)]
Rollout 89, Act: 2.762
6 T = 100. So T = 100 /0.6 ��
Rollout 10, Act: 2.731
considering that between consecutive squares (k2 and (k+1)2), the possible moves that can
Rollout 82, Act: 2.692
must be such that groups with higher total probability have shorter average binary codeword lengths.\n\nBut actually,
Rollout 82, Act: 2.688
binary subtree. So, assigning more probable symbols to shorter codes in their respective subtrees.\n\nWait, the
Rollout 18, Act: 2.679
:\n\n5 y =x c +4 => y = (x c +4)/5\n\nSet equal
Rollout 11, Act: 2.671
+ y2 )=0 → y [1 -1/(x2 + y2 ) ]=
Rollout 10, Act: 2.665
W if there exists a square number s <=n such that n - s is L.\n\nn is L
Top 16 Negative Activations
Rollout 4, Act: -2.775
approximately 0.36%, consistent with the previous result.\n\nTherefore, combining this with the 2
Rollout 91, Act: -2.778
2?\n\nBut wait, wait. So according to my computations in this coordinate system, the two terms (
Rollout 3, Act: -2.791
in kinetic theory, they are different. Let me make sure.\n\nAccording to Maxwell-Boltzmann distribution
Rollout 36, Act: -2.798
to attempt factoring the quartic equation.\n\nSo current equation is:\n\ns^4 -30,
Rollout 30, Act: -2.807
n+1} b_n -1. So our numerator is\n\na_{n+1} b
Rollout 81, Act: -2.811
. Now, is there any possible miscalculation here?\n\nAlternatively, let me think of another approach
Rollout 73, Act: -2.814
from k=1 to n, so subtract the k=0 term (which is 0), so
Rollout 20, Act: -2.843
43, p^{kn} �� 1 mod 43. Similarly, for modulus
Rollout 80, Act: -2.867
to calculate the total number of neutrinos passing through the core per second. But wait, the flux density
Rollout 30, Act: -2.904
n+1 } is equal to:\n\nFirst term when n=1: 2013 /
Rollout 15, Act: -2.918
not cycling. Therefore, my mistake was introducing that previously. So maybe n=3 isn't cyclic.\n\n
Rollout 43, Act: -2.952
>.\n\nI need to be careful here. So in my calculation, coefficients matter. But let me redo once
Rollout 30, Act: -3.064
's adjust indices for these sums. Let k = n -1 for the first sum and m = n
Rollout 97, Act: -3.101
~2.3.\n\nBut then why was the previous answer approx 4.6?\n\nBecause, in
Rollout 84, Act: -3.234
twice the fraction, or twice the volume.\n\nIn my initial interpretation, it's twice the volume. That
Rollout 69, Act: -3.436
).\n\nSo, for S(theta), first term is i=0: sin2(theta)/cos(3

Layer 32

GATE_PROJ

Top 16 Positive Activations
Rollout 68, Act: 1.019
quadratic equation has two real solutions. However, we need to check if these solutions lie within the domain constraints
Rollout 70, Act: 0.871
we verify \( 2 \times 315 = 630 \).\n\nThus, the
Rollout 49, Act: 0.853
rizable, consider a Bernstein set. A Bernstein set is uncountable, meets every perfect set,
Rollout 49, Act: 0.846
?\n\nWait, maybe the Bernstein set. A Bernstein set is a subset B of a Polish space such that
Rollout 68, Act: 0.808
the quadratic equation has two real solutions. However, we need to check if these solutions lie within the domain
Rollout 72, Act: 0.767
: \( e^{2\pi i k/5} \) for k=1,2,
Rollout 93, Act: 0.764
corresponds to angles that are π/1007, 2π/1007,
Rollout 72, Act: 0.763
n=5 are 1 and the primitive 5th roots: \( e^{2\pi i
Rollout 68, Act: 0.753
+ 4 = 0\n\]\n\nWe need this quadratic equation to have exactly one real solution in
Rollout 33, Act: 0.738
\(1 \leq k \leq 5\), at least one of the first \(k
Rollout 83, Act: 0.732
agon, so every set of three consecutive vertices wraps around.\n\nFirst, maybe I should think about the possible
Rollout 6, Act: 0.724
\) unless \( \textbf{NP} = \textbf{P} \).\n\n---\n\n**
Rollout 2, Act: 0.695
91 mod125: 125*7=875. 89
Rollout 90, Act: 0.682
:\n\n1,928 is less than2018. Therefore,10,00
Rollout 93, Act: 0.681
}^{N-1} e^{i2πk m/N}=N if m ��0
Rollout 72, Act: 0.670
=5. The roots of unity for n=5 are 1 and the primitive 5th roots
Top 16 Negative Activations
Rollout 68, Act: -3.683
from -500 to 500 is 500 negative integers (k=-1
Rollout 70, Act: -3.687
7^0.\n\nBut if we take those digits d_k d_{k-1} ... d_
Rollout 38, Act: -3.688
and W south, so the distance is sqrt(x2 + W2). Therefore, his time is sqrt
Rollout 83, Act: -3.693
same residue. Remember, residue counts: 0,1,2 each needs to be assigned exactly three
Rollout 61, Act: -3.696
of m's: for d, it's m from 0 to 39 -d, hence
Rollout 35, Act: -3.709
0.\n\nLet’s test D =1: b2 +7b -1=0. Discriminant
Rollout 22, Act: -3.724
2 * 1.6605 is about 3.321. Then 0
Rollout 71, Act: -3.748
multiple possible m, but we just need the sum |a| + |b| + |c|
Rollout 10, Act: -3.774
If n ��0 mod4: subtract0 or1.\n\n subtract s≡0 or1 mod4
Rollout 31, Act: -3.779
1. Indeed, there would be coefficients c_n and d_n such that a_n = c_n a
Rollout 4, Act: -3.795
significant figures, so 0.88%, or maybe the answer is exact at 0.8
Rollout 0, Act: -3.835
1.\n\nSo primes 2: 18, 3: 8, 5:4
Rollout 91, Act: -3.836
AD is tan(45) = 1 or tan(135) = -1.
Rollout 53, Act: -3.862
3)) ) Rationallze: this is or we can leave as it is. Maybe we can
Rollout 72, Act: -3.927
. Then we want to solve for exponents i, j, k such that ��^i +
Rollout 26, Act: -4.244
J = 296.0 kJ but only three sig figs. So 29

UP_PROJ

Top 16 Positive Activations
Rollout 80, Act: 2.663
Earth's surface is 10^11 neutrinos per second per square centimeter. \n\nThe
Rollout 80, Act: 2.656
first, we need to calculate the total number of neutrinos passing through the core per second. But wait
Rollout 80, Act: 2.416
is), then the neutrinos are coming in parallel rays. The number passing through the inner core's cross
Rollout 80, Act: 2.378
neutrino flux: 10^11 neutrinos/s/cm2. How many neutrinos hit
Rollout 74, Act: 2.371
Let’s define:\n\n- H_e: Earth's horizontal magnetic field towards north.\n\n- H_ship_N:
Rollout 80, Act: 2.354
surface.\n\nTherefore, to calculate the total number of neutrinos passing through the Earth per second, we could
Rollout 74, Act: 2.351
down the components.\n\nAssume that the Earth's horizontal magnetic field is H_e. The ship's magnetic
Rollout 74, Act: 2.309
components.\n\nAssume that the Earth's horizontal magnetic field is H_e. The ship's magnetic fields are
Rollout 74, Act: 2.284
The compass needle points in the direction of the resultant horizontal magnetic field. The Earth's field is H_e
Rollout 80, Act: 2.280
, we need to calculate the total number of neutrinos passing through the core per second. But wait,
Rollout 80, Act: 2.266
Earth's inner core", which probably means all the neutrinos that hit the Earth's surface also are absorbed
Rollout 80, Act: 2.250
determines the total power.\n\nTherefore, the number of neutrinos absorbed per second would be the flux (1
Rollout 74, Act: 2.232
H_e and H_s = K_s * H_e, then tanδθ = [K_b H
Rollout 80, Act: 2.220
11 neutrinos/s/cm2. How many neutrinos hit the core? Assuming the core is a
Rollout 80, Act: 2.184
the center, but the cross-sectional area that the neutrinos pass through would be the cross-section of the
Rollout 80, Act: 2.177
surface, the flux is 1e11 neutrinos/s/cm2. Because neutrinos travel in
Top 16 Negative Activations
Rollout 73, Act: -2.363
$ is divided by $1000$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I
Rollout 66, Act: -2.364
If m and n are coprime and opposite parity, then let me consider:\n\ngcd( m2
Rollout 72, Act: -2.364
suppose that one of them is 1 (i=0). Then the equation would be 1 +
Rollout 11, Act: -2.366
ate of variables. Let’s set x = \overline{a}, y = \overline{
Rollout 41, Act: -2.368
both digits are non-zero. Then, we can arrange those two digits freely, but ensuring no three of
Rollout 66, Act: -2.372
2 + n2\n\nSo, let's compute all terms:\n\nFirst, a = m2 - n
Rollout 66, Act: -2.386
here. Therefore, m and n must have opposite parity. Hence, the parameterization requires that in a
Rollout 11, Act: -2.395
only.\n\nAlternatively, maybe take conjugate of variables. Let’s set x = \overline{a
Rollout 66, Act: -2.406
ence, m and n have to be of opposite parity. If m=13 is odd, n
Rollout 71, Act: -2.409
signs accordingly, maybe. For example, swapping e and d, but since coefficient is 3, if
Rollout 11, Act: -2.437
's denote x = a, y = b, z = c. Then, given xyz = 1
Rollout 71, Act: -2.441
have all even.\n\nSimilarly, two even and one odd: Let two be even, one odd. Then
Rollout 61, Act: -2.457
2^k + 2^m �� 0 mod 9. Let’s write this as
Rollout 11, Act: -2.496
:\n\nGiven abc=1, then taking reciprocals, (1/a)(1/b)(1/c
Rollout 66, Act: -2.532
for some m > n coprime integers of opposite parity. Or maybe they could be in some other
Rollout 41, Act: -2.633
4=504.\n\nCase B: One of the digits is zero. Here, thousands digit can

DOWN_PROJ

Top 16 Positive Activations
Rollout 22, Act: 2.974
000,000=12,666\n\n0.0063
Rollout 9, Act: 2.902
celes triangle with base EM: if it's vertical, then the base is vertical. But triangles are
Rollout 42, Act: 2.878
a cyclopropane carbon, which originally had three bonds (as part of the cyclopropane
Rollout 97, Act: 2.831
45:\n\n1.45=1.5 -0.5 e^{-kt} →
Rollout 72, Act: 2.831
). Cos(36°)= (sqrt(5) + 1)/4 * 2?
Rollout 90, Act: 2.800
14126. Middle digit is the third digit, which is 1. Removing gives B
Rollout 42, Act: 2.758
lopropyl group. So the central carbon has three substituents: one hydroxyl (from
Rollout 42, Act: 2.719
a four-membered ring attached via one of its carbons to the central carbon), a cycloprop
Rollout 54, Act: 2.694
in SN2, a bulky nucleophile (like tert-butoxide) is less nucleophilic
Rollout 9, Act: 2.681
triangle GEM with base EM. So EM is one of the sides of the square, the bottom side
Rollout 72, Act: 2.681
72 degrees) is equal to (sqrt(5) - 1)/4 * 2.
Rollout 9, Act: 2.673
triangle. The triangle is isosceles with base EM, so the legs would be GE and GM
Rollout 42, Act: 2.642
the spiro carbon), and the double bond at position 5. Wait, the numbering would have the
Rollout 42, Act: 2.639
carbocation is at the central carbon (tertiary maybe?), perhaps adjacent to a three-membered ring
Rollout 9, Act: 2.634
apex of the triangle opposite the base EM, which is the vertical side of the square. If the altitude
Rollout 72, Act: 2.633
) is known to be (1 + sqrt(5))/4 multiplied by 2. Perhaps a better
Top 16 Negative Activations
Rollout 30, Act: -2.433
n+1} to a_{n-1} and b_n.\n\nSo to express a_{m
Rollout 18, Act: -2.435
are integers, then a x + b y = s x + t y where s = a and t
Rollout 50, Act: -2.446
First, recall that \(\log(a \cdot b) = \log a + \log b\
Rollout 81, Act: -2.458
20h/(51 -h2) =22/7 but in this approach, by
Rollout 81, Act: -2.460
Not right-angled, But we can make decomposition via altitude.\n\nConsider triangle ABC altitude ha =h leading
Rollout 49, Act: -2.479
orff space is regular. Let's see: Take a point x and a closed set C not containing
Rollout 56, Act: -2.492
that her pasture never exceeds an area of $K$?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so
Rollout 18, Act: -2.494
4). To check if H is Z2, we can compute the determinant of the matrix formed by two
Rollout 18, Act: -2.494
of all integer linear combinations of (0, x) and (1, y). That is, every
Rollout 18, Act: -2.548
_{xy} is generated by (0, x) and (1, y), then elements of H
Rollout 38, Act: -2.562
. So, we have Ana, Bob, and Cao biking at different speeds: 8.6
Rollout 1, Act: -2.573
l is non-empty if and only if r1 + r2 >= l and |r1 - r
Rollout 18, Act: -2.616
2. Therefore, H is isomorphic to Z2 if the generators are linearly independent over Z
Rollout 11, Act: -2.629
Hmm, complex numbers with product 1. So abc = 1. None of them are real or
Rollout 56, Act: -2.633
go to infinity. The question is essentially asking for the sum of the series of areas, if it conver
Rollout 58, Act: -2.659
of k and set up the equation that one is 2018 times the other.\n\nLet me

Layer 33

GATE_PROJ

Top 16 Positive Activations
Rollout 62, Act: 4.912
17303. Let's see: 121*100=12
Rollout 81, Act: 4.879
). Calculate sqrt(25909): closest square is 1612=25
Rollout 36, Act: 4.877
,662,592):\n\nSince 19,0002 = 3
Rollout 36, Act: 4.832
imates to 158.12 because 1582=24,96
Rollout 96, Act: 4.797
�13:\n\n13*133 = 1729, so 17
Rollout 3, Act: 4.706
8,436). Let's see, 168^2 = 28,
Rollout 81, Act: 4.674
9584) = let's check: 172^2=2958
Rollout 81, Act: 4.651
=25,921. ,Since 1612=25,92
Rollout 16, Act: 4.621
80 divided by 1953: 1953*4=781
Rollout 96, Act: 4.570
by2197:\n\n2197*31=2197*30
Rollout 38, Act: 4.527
27*527=?\n\nWell, 5002=250,0
Rollout 3, Act: 4.509
8,408). Let's see.\n\n168^2 = (170 -
Rollout 36, Act: 4.509
1,856= let's check:\n\n10,000*10,0
Rollout 90, Act: 4.495
8073 ��1009: 1009*8=80
Rollout 53, Act: 4.454
1 ��146:\n\n146*13=1898 → 1
Rollout 81, Act: 4.452
29584) = let's check: 172^2=295
Top 16 Negative Activations
Rollout 80, Act: -4.138
pass through the Earth, and the flux is given at Earth's surface, but only those that pass through
Rollout 27, Act: -4.163
T = (m v_rms2)/(3k). Then once we know m (mass of Be
Rollout 25, Act: -4.164
e6\n\nBut a_earth = G*m / d2(t), depends on time-varying
Rollout 97, Act: -4.168
is written as:\n\nRate = - (1/a) d[A]/dt = k [A]^n
Rollout 27, Act: -4.174
v_z2⟩ = (1/2)k T → ⟨v_z2⟩ = k T
Rollout 27, Act: -4.175
= (2 ΔE_rms) / (3k). Get 2*5.54
Rollout 27, Act: -4.182
T ⇒ T = 2ΔE_rms /k.\n\nOnce we have T, but since the
Rollout 27, Act: -4.187
lead to T = 2 ΔE_rms / k.\n\nThen, once T is known, T
Rollout 27, Act: -4.190
T = (2 * delta E_rms) / k.\n\nPlugging numbers:\n\ndelta E_rms
Rollout 25, Act: -4.227
t2 = (1/2)*( G*m /d_min2 ) * (d_min/v )
Rollout 80, Act: -4.354
However, if the neutrinos are coming uniformly from the Sun, then their paths are parallel, all coming
Rollout 27, Act: -4.384
ΔE_rms = (1/2)kT ⇒ T = 2ΔE_rms
Rollout 27, Act: -4.405
thus T = m V_rms2/(3k). And they call that Tc. Perhaps confused
Rollout 12, Act: -4.428
1 -k2)K(k) ] / (k(1 -k2)) * ( -
Rollout 80, Act: -4.452
. However, if the neutrinos are coming uniformly from the Sun, then their paths are parallel, all
Rollout 12, Act: -4.585
(1 -k2)K(k) ] / (k(1 -k2)) * (

UP_PROJ

Top 16 Positive Activations
Rollout 36, Act: 3.205
,760≈2.016\n\nThus, approximate sqrt(Δ)=19,
Rollout 70, Act: 3.054
8 =>d1 <=2.666, so d1=0,1,2.\n\n
Rollout 23, Act: 2.932
91≈0.05126. So that 73.0422
Rollout 4, Act: 2.922
0036058 as a decimal, so yes, 0.3605
Rollout 68, Act: 2.894
6)/2≈-0.008. So x≈-0.008,
Rollout 23, Act: 2.868
2585≈0.722. So, approximately 31.722
Rollout 3, Act: 2.777
)/337≈0.047. Thus, sqrt≈168.5 +
Rollout 17, Act: 2.733
*0.41=3.28, so total 11.28.\n\nThus
Rollout 13, Act: 2.701
1.12550881, therefore 1.12550
Rollout 23, Act: 2.693
72.991≈0.05126. So that 73.
Rollout 22, Act: 2.677
=145, 659. Remainder is 145,458
Rollout 65, Act: 2.673
k*c -> k=1995/c, and from c = k*1995
Rollout 22, Act: 2.672
2*273=. Hmm, 0.5*273=136.
Rollout 26, Act: 2.669
66= 2.4734\n\nThen multiply by -3: -3 *
Rollout 36, Act: 2.651
400=78,192\n\nSo => dk≈78,192
Rollout 36, Act: 2.635
; =>x≈2.0105...\n\nBut likely, sqrt(36,86
Top 16 Negative Activations
Rollout 70, Act: -2.566
\( d_k d_{k-1} \ldots d_0 \) in base-7
Rollout 46, Act: -2.568
+ (1/2)ω2(x2 + y2) + �� (x dy/d
Rollout 7, Act: -2.568
59 for non-promoted participants. \n\n### Part (a)\n\n1. **Original equations:
Rollout 32, Act: -2.583
\omega \\ 1 & 0 \end{pmatrix}\)\n2. \(\mathbf
Rollout 35, Act: -2.588
\) and \( P(24) = 17 \), we find:\n\[\nQ
Rollout 91, Act: -2.591
(t - 1, t)\) and \(\overrightarrow{CB} = (-2,
Rollout 46, Act: -2.591
2 - (ω2/2)(x2 + y2). Hmm, right. Let me try
Rollout 0, Act: -2.608
) and \(\gcd(a, b) = 1\), each prime factor of 20
Rollout 53, Act: -2.616
= 2\) implies \(AD = 3 \times 2 = 6\).\n\n2.
Rollout 46, Act: -2.627
^2}{2}\left(x^{2} + y^{2}\right)\n \]\n\n2
Rollout 6, Act: -2.635
sets \( S_1, S_2, \ldots, S_k \) by tagging each
Rollout 25, Act: -2.635
in a closest approach distance. In reality, 2023 BU's closest approach was 3
Rollout 5, Act: -2.636
\) of the segment that divides the trapezoid into two regions of equal area, we use
Rollout 5, Act: -2.679
00 \). The midline of the trapezoid, which connects the midpoints of the
Rollout 97, Act: -2.707
pressure \(P_0 = 1 \, \text{atm}\), final pressure \(P
Rollout 61, Act: -2.752
1 \pmod{9} \). This simplifies to finding \( d \) such that \(

DOWN_PROJ

Top 16 Positive Activations
Rollout 35, Act: 3.358
(b +7)*b must be a divisor of 10. Therefore, we need integer values
Rollout 56, Act: 3.256
straight line. So A0 is between A2 and A3. So the length A3A0
Rollout 62, Act: 3.185
561÷29=88.31, no. 31:31
Rollout 67, Act: 2.963
)= -24/k. Therefore, k must divide 24.\n\nSo k is an integer that
Rollout 65, Act: 2.909
2)*1995, so n2 divides 1995*m2. But since
Rollout 71, Act: 2.902
96. Close, but 85.81, not. c=1:3*
Rollout 56, Act: 2.888
3 is a straight line. So A0 is between A2 and A3. So the length A
Rollout 83, Act: 2.864
(9, k) = d, so d divides 9. Then, rotation by k will split
Rollout 3, Act: 2.830
28,224.\n\n168.5^2= (168 +
Rollout 67, Act: 2.813
for p and q to be integers, k divides into 96 and 36. The gcd
Rollout 67, Act: 2.775
15d/2.\n\nBut a must divide into 1440 as per equation 1
Rollout 67, Act: 2.775
So 6a2 +29 a d -120 d2=0.\n\nAn equation
Rollout 35, Act: 2.768
(b +7)*b must be a divisor of 10, because c has to be integer
Rollout 71, Act: 2.698
011/7=287.285... 11: 20
Rollout 32, Act: 2.695
0ω\n\nTerm3 has +670 and -670ω2\n\nTerm4 has
Rollout 66, Act: 2.691
6b2 - 6b c = 0\n\nHmm, perhaps group terms:\n\na2 -
Top 16 Negative Activations
Rollout 55, Act: -2.514
in the x-z plane and a 1x1 reflection matrix in the y-axis. The 2
Rollout 63, Act: -2.517
- \( f(1, 1) = 1 \)\n- \( f(2,
Rollout 62, Act: -2.519
121, which is 112. Then m=12: 14
Rollout 75, Act: -2.520
a sum of two squares. For example, the square of an odd integer is 1 mod 4
Rollout 61, Act: -2.533
8 mod 9, where d = k - m. Because -1 mod 9 is
Rollout 55, Act: -2.539
2x2 rotation matrix and the 1x1 matrix [-1], then when we raise A to
Rollout 15, Act: -2.560
's try n=3, which is 2^2 -1: no, since n=3
Rollout 62, Act: -2.651
11 yields 121=112. But (11k)^2 -1
Rollout 55, Act: -2.665
Thus, if A is a block diagonal matrix with blocks: the 2x2 rotation matrix and the
Rollout 95, Act: -2.695
-1} and PkQk is the average of the lengths of the two segments multiplied by the
Rollout 17, Act: -2.712
is unitless). Wait, �� has units of 1/m^(3/2), so |
Rollout 5, Act: -2.819
bases that divides the trapezoid into two regions of equal area, its length x should be such
Rollout 3, Act: -2.883
is a gas at standard temperature and pressure, with boiling point at -61.7 °C,
Rollout 15, Act: -2.894
1? Wait, n=1 is 2^1 -1. Let's try n=3
Rollout 55, Act: -2.932
in the x-z plane and a 1x1 matrix (-1) acting on the y-axis.
Rollout 72, Act: -3.069
check n=5. The roots of unity for n=5 are 1 and the primitive 5

Layer 34

GATE_PROJ

Top 16 Positive Activations
Rollout 38, Act: 4.108
2 -625r +168=0.\n\nUsing quadratic formula:\n\nr=(62
Rollout 53, Act: 3.900
in the ratio 2:1. So CP:PE = 2:1. Since PE is
Rollout 96, Act: 3.821
1=5α -12β\n\nSolving these numerically:\n\nFrom first equation, -2
Rollout 96, Act: 3.807
697*895\n\nAdding them to eliminate β:\n\n(2316*2
Rollout 61, Act: 3.794
. Check 17: 133 divided by 7 is 19; yep
Rollout 28, Act: 3.790
4a2y2 +8ab xy.\n\nCompare to our equation:\n\n4b2x2+
Rollout 53, Act: 3.786
along the median from A to D. So AP:PD = 2:1. PD is given
Rollout 9, Act: 3.770
is 100 -500/h=80. Then 500/h=
Rollout 81, Act: 3.759
- 561 = 0\n\nLet's attempt to solve this quadratic equation for h. Quadr
Rollout 69, Act: 3.753
4 sin^2 theta)\n\nSo, let's substitute this into term_i:\n\nsin2(theta_i
Rollout 9, Act: 3.747
(10,0):\n\nUsing point-slope form: y -0 = (-5/h)(x
Rollout 81, Act: 3.723
-140h -1122=0.\n\nBut before, it was 22
Rollout 86, Act: 3.719
2. That's straightforward: [x^5 /5] from 0 to pi/2 =
Rollout 23, Act: 3.712
Possibly considering Bateman equation or parent-daughter, but that doesn't apply here as Y and Q are
Rollout 62, Act: 3.711
55. Not. 2561 divided by 17: 17*1
Rollout 7, Act: 3.707
P = 56N +15P=66N\n\nSo 15P=
Top 16 Negative Activations
Rollout 71, Act: -2.179
But 3c2 is even if c is even and odd otherwise. Similar for k2. Since
Rollout 33, Act: -2.191
there should be at least one number greater than 2, which means they can't both be 1
Rollout 75, Act: -2.194
universally.\n\nWait.\n\nActually, if we have three even numbers.\n\nWait, time to pause. Let me
Rollout 19, Act: -2.201
by10. Remainder5.\n\nn=6: sum=441 divided by11
Rollout 33, Act: -2.209
can't be 1. Then for k=2, in the first two elements, there should be
Rollout 2, Act: -2.210
^1 -1), the second is k=2 (10^2 -1), ..., up
Rollout 75, Act: -2.214
sum of the squares of any three pairwise different positive odd integers can be represented as the sum of the squares
Rollout 75, Act: -2.217
So we need three of the six squares to be odd (1 mod 4) and three to be
Rollout 75, Act: -2.230
sum of the squares of any three pairwise different positive odd integers can be represented as the sum of the squares
Rollout 83, Act: -2.267
- Residue 0: {3, 6, 9}\n- Residue 1:
Rollout 75, Act: -2.305
sum of the squares of any three pairwise different positive odd integers can be represented as the sum of the squares
Rollout 75, Act: -2.306
(1 mod 4) and three to be even (0 mod 4). Therefore, in order
Rollout 93, Act: -2.367
from1 to1007. For k even, T_{1007}(x_k
Rollout 58, Act: -2.406
to k of (i/(i+1))^i.\n\nSimilarly, P(1) is the probability
Rollout 33, Act: -2.492
k$ terms of the permutation is greater than $k$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay,
Rollout 75, Act: -2.613
eligible?\n\nWait, consider even numbers squared:\n\nEach even square is a multiple of4.\n\nTherefore, sum

UP_PROJ

Top 16 Positive Activations
Rollout 25, Act: 1.867
=5.97e24 kg, then m≈1.027e25
Rollout 60, Act: 1.746
_observer is the speed of the observer (car) towards the source.\n\nBut actually, the formula depends on
Rollout 47, Act: 1.698
�� v_air2\n\nThus, the pressure difference p0 - p = (1/2)
Rollout 96, Act: 1.589
169))/5\n\nLet's compute numerator:\n\n90 = (90*169
Rollout 30, Act: 1.583
the initial conditions, which is a good sign.\n\nTherefore, this fraction is indeed the answer.\n\nBut is
Rollout 99, Act: 1.572
f(x1), f(x2), compute the differences:\n\nh0 = x1 - x0\n\n
Rollout 8, Act: 1.514
original problem, seems that my answer is right.\n\nTherefore, probability is 31/126
Rollout 64, Act: 1.498
then k=0, so B=(a2 -a1)/2=0, and A=
Rollout 77, Act: 1.471
. So the answer is 364.\n\nTherefore, the function f is actually the least common multiple
Rollout 99, Act: 1.468
= (f(x3) - f(x2))/h1 = (1.1765
Rollout 45, Act: 1.453
(-1) = -9 +b, so:\n\nf(-2) = 2(-9 +
Rollout 88, Act: 1.451
. So the answer is 100. Therefore, I think it's 100.
Rollout 26, Act: 1.442
122.64 kJ.\n\nTherefore, the reaction as written (using Hg(g
Rollout 30, Act: 1.434
4 )\n\nHence, the result is correct.\n\nTherefore, 1 +1/(2015
Rollout 44, Act: 1.433
*V1), so nRT ln(V1/V2) = p1*V1 ln(V
Rollout 73, Act: 1.413
is 0), so it's the same.\n\nSo sum_{k=1}^{n}k
Top 16 Negative Activations
Rollout 86, Act: -2.558
2. That's straightforward: [x^5 /5] from 0 to pi/2 =
Rollout 1, Act: -2.561
of them.\n\nAlternatively, for the system to have a solution, then for any x, x' in
Rollout 18, Act: -2.576
3,8), (4,-1), and (5,4) is the same as the subgroup
Rollout 15, Act: -2.599
lamps will eventually turn off after some time.\n\nFirst, let me rephrase the rules to make sure I
Rollout 45, Act: -2.600
4m for all non-negative integers x. Find the greatest possible value of m.\n\nWait, but f
Rollout 24, Act: -2.607
an associated system of inequalities that determine R and C, which are structurally similar across different minimal pairs.
Rollout 1, Act: -2.610
with equal norms and their differences have equal norms, this often relates to them being vertices of a simplex or
Rollout 36, Act: -2.613
equations with variables x, y, z, and s. We need to solve these equations to find x
Rollout 85, Act: -2.622
,y)= a x + b y, with a and b being 1 and 2 in some order
Rollout 18, Act: -2.645
=H_{xy}, we need to determine x and y such that:\n\n1. The generators of H
Rollout 51, Act: -2.645
π:\n\n∫0^π cosnx dx = [sin(nx)/n]0^π = (
Rollout 37, Act: -2.668
inherent symmetry that groups these terms into multiplets which a commonality leading to divisibility. For example,
Rollout 1, Act: -2.729
orthogonal to the set S - S, except for a constant.\n\nBut in that case, if we take
Rollout 51, Act: -2.737
x) times the absolute value of sin(x - ��) dx, where �� is between
Rollout 8, Act: -2.785
available.\n\nTherefore, this is similar to permutation with two colors. The number of such sequences is C(
Rollout 14, Act: -2.796
*1? Not sure either.\n\nAlternatively, use homogeneous functions. Suppose f is a scaling function such that

DOWN_PROJ

Top 16 Positive Activations
Rollout 87, Act: 2.914
/ sqrt(η2 + ��2)) evaluated from 0 to 1.\n\n= sinh^{-
Rollout 86, Act: 2.899
Thus, integral = -x^3 cos x | from 0 to pi/2 + 3
Rollout 87, Act: 2.874
1(z / sqrt(x2 + y2)) from 0 to a\n\nWhich is equals sinh^{-
Rollout 86, Act: 2.805
pi/2 is (1/2)x2 evaluated from 0 to pi/2, which is
Rollout 14, Act: 2.772
( \int f(x)^2 dx - C \right) - \mu \left( \int
Rollout 14, Act: 2.761
integral:\n\n∫ x dx = x2 /2 from 0 to 1/μ: (1
Rollout 86, Act: 2.721
x2)(1/2) sin2x | from0 to pi/2 - �� (
Rollout 86, Act: 2.716
(1/2)x^2 sin2x | - ��x sin2x dx =0
Rollout 51, Act: 2.703
x/9 + cos7x/7] evaluated from 0 to π.\n\nCompute at π:\n\n
Rollout 86, Act: 2.643
: (1/2)x2 sin2x | from0 to pi/2 =0.\n\nThen
Rollout 9, Act: 2.517
10 -100/h)*x ] from 0 to10.\n\nEvaluating at
Rollout 51, Act: 2.475
to pi is [-cos9x/9] from 0 to pi which is (-cos9pi
Rollout 86, Act: 2.364
x.\n\nThus, integral = x3 sin x | from 0 to pi/2 - 3
Rollout 86, Act: 2.361
.\n\nThus, 3 [ x2 sin x | from0 to pi/2 - ��
Rollout 86, Act: 2.359
x dx from0 to pi/2=sinx |0 to pi/2=1 -0=
Rollout 86, Act: 2.330
Integral= (x/2) sin2x |0 to pi/2 - (1/2
Top 16 Negative Activations
Rollout 78, Act: -2.612
9/800) = (22/3) * (1/800)
Rollout 14, Act: -2.615
? Since Hölder can relate multiple integrals with exponents that are conjugate. Let me recall Hö
Rollout 55, Act: -2.659
rotation block would be rotated by 2*(π/6) = π/3. A^3
Rollout 14, Act: -2.670
/2} dx.\n\nApply Hölder with exponents p=2, q=4, and r
Rollout 14, Act: -2.680
's inequality for three functions would state that for exponents $p, q, r$ such that $
Rollout 96, Act: -2.706
90 = (90*169)/169 = 15210
Rollout 14, Act: -2.734
.\n\nLet’s remember that Hölder’s inequality with exponents 2 and 2 gives Cauchy
Rollout 53, Act: -2.737
3 ) /2 + (-27/(2 sqrt73 )) = same as 73/(
Rollout 14, Act: -2.752
3/2? Wait, Hölder's with exponents p, q such that 1/p +
Rollout 14, Act: -2.776
} dx.\n\nThen apply Hölder's inequality with exponents p, q, r such that 1
Rollout 14, Act: -2.823
Let’s remember that Hölder’s inequality with exponents 2 and 2 gives Cauchy-S
Rollout 14, Act: -2.835
dx.\n\nThen apply Hölder's inequality with exponents p, q, r such that 1/p
Rollout 36, Act: -2.837
*(s2 + 8064)/(2s) = - (s2 + 8
Rollout 53, Act: -2.853
sqrt73 /2 -27/(2 sqrt73 ) = (73 -27
Rollout 14, Act: -2.856
lder's inequality for three functions would state that for exponents $p, q, r$ such that
Rollout 14, Act: -2.972
three functions f, g, h, and exponents p, q, r with 1/p +

Layer 35

GATE_PROJ

Top 16 Positive Activations
Rollout 26, Act: 4.381
0 has four sig figs, 312.8 has four, their difference is 1
Rollout 68, Act: 4.226
≈[ -504 + 503.98413 ] /2≈
Rollout 68, Act: 4.167
/2≈(-504 +503.984)/2≈(-0.
Rollout 17, Act: 4.137
T, which would be approximately 25.4 mT, so 0.025
Rollout 81, Act: 4.047
2601*(533-49)=0.\n\n Compute:\n\n533-4
Rollout 53, Act: 4.034
be answer.\n\nHence, the area is \boxed{\dfrac{27}{2}}.\n\n
Rollout 0, Act: 4.028
answer is 128, corresponding to \boxed{128}? Or wait, wait
Rollout 68, Act: 4.021
[ -504 + approx 504 ] /2≈ something like (-504
Rollout 68, Act: 4.021
would be [ -504 -503.98413]/2≈[-
Rollout 4, Act: 4.009
2 + (2 * 0.004)^2 )\n\nCalculating 2*0.
Rollout 81, Act: 3.994
2601*(533 -49)=0.\n\nCompute coefficients:\n\nCompute first coefficient:
Rollout 40, Act: 3.981
steps check out. Therefore, the answer is \boxed{19}.\n\n**Final Answer**\n\
Rollout 45, Act: 3.979
1 +4= [(-1 +4 -4)]*2 + (-6) +8 +
Rollout 19, Act: 3.976
is78. Therefore, box it as \boxed{78}. But let me try k=
Rollout 32, Act: 3.968
18. Therefore, the answer is boxed \boxed{4018}.\n\nBut before final
Rollout 4, Act: 3.951
90=4975*(1200-10)=4975*
Top 16 Negative Activations
Rollout 87, Act: -2.712
dξ dη dζ\n\nTo compute K, we can look for possible solutions. I might have
Rollout 25, Act: -2.728
m )/v2\n\nSolving for m,\n\nm= (3.541e6
Rollout 1, Act: -2.746
||y||2 -d2/2.\n\nTherefore, each inner product of xi with y is this
Rollout 13, Act: -2.758
to multiply directly. Still, this might take time.\n\nAlternatively, using currency calculation steps:\n\n1.5
Rollout 14, Act: -2.772
λ), and zero elsewhere.\n\nCompute the integrals:\n\nJ = ��0^{x_max} [(
Rollout 31, Act: -2.782
} -d_{m+3}=0.\n\nTherefore, this gives us linear equations for the coefficients c
Rollout 36, Act: -2.788
x + 12600/s\n\nTherefore:\n\ns2 - s(y + z) =
Rollout 51, Act: -2.798
2π cos8x |sinx| dx\n\nNow, considering the integral over 0 to
Rollout 87, Act: -2.814
atively. Let me try to compute K as follows:\n\nK = ��0^1 ��0
Rollout 87, Act: -2.839
1/r over a cubic volume. That seems challenging.\n\nAlternatively, is there a known integral for potential at
Rollout 30, Act: -2.863
n+1} -1 ) -1.\n\nTherefore, the term we have in the sum:\n\n[
Rollout 87, Act: -2.888
2) dx. Let me compute this integral first.\n\nLet me define f(y,z) = ��
Rollout 4, Act: -2.897
\n\nHmm, let's compute these step by step.\n\nFirst, compute the coefficients for ΔC_eq.\n\n
Rollout 14, Act: -2.900
1 - μ x)/(2 λ)] dx.\n\nSo first, let's compute J:\n\nLet’s do
Rollout 73, Act: -3.004
, perhaps not. Maybe I need a better approach.\n\nAlternatively, let's compute S = sum_{j
Rollout 89, Act: -3.100
/5)T = 100.\n\nSo T = 100 * (5/

UP_PROJ

Top 16 Positive Activations
Rollout 7, Act: 2.641
?\n\nWait, wait:\n\nWait 18N is equal to 54N/3. So
Rollout 26, Act: 2.477
kJ/K (4 sig figs)\n\n9- ΔT=14.8 K (3
Rollout 28, Act: 2.425
, I remember that a conic is defined by five points, but here we have four points: O
Rollout 30, Act: 2.410
*2014=4,058,210. Therefore, the total sum
Rollout 75, Act: 2.375
the problem says six squares. So above approach gives seven squares, which is more than allowed. So need
Rollout 32, Act: 2.351
��3 = 1, so ��4 = ��. Therefore, ��4 -
Rollout 70, Act: 2.283
but we had N=7k +d0, and d0=2. Thus, N=
Rollout 17, Act: 2.281
Wait, we canceled out the 10^-33 in numerator and denominator, so the exponent is
Rollout 30, Act: 2.238
5*2014=4,058,210. Therefore, the total
Rollout 1, Act: 2.233
2 = 3d2/4 - which is problematic.\n\nHowever, at the same time,
Rollout 64, Act: 2.224
}-1 ) )\n\nBut 2^{2012}=8*2^{2009
Rollout 6, Act: 2.190
are in $\textbf{P}_{angel}$ because the \textit{angel string}
Rollout 64, Act: 2.178
so we can substitute: 2^{2012} =8*2^{200
Rollout 22, Act: 2.172
.\n\nR is the gas constant, which is 8.314462618
Rollout 75, Act: 2.162
squares and 3 ones. Which is seven squares. No.\n\nWait, the problem says six squares.
Rollout 68, Act: 2.153
*508)=sqrt(254000). Hmmm.\n\nBut regardless, the
Top 16 Negative Activations
Rollout 0, Act: -2.342
a}{b}\) where \(0 < a < b\) and \(a\) and \(b\)
Rollout 28, Act: -2.342
0):\n\nab = -2k a2 / (2h -a) -2h b2
Rollout 9, Act: -2.364
. However, an isosceles triangle with base EM: if it's vertical, then the base
Rollout 65, Act: -2.366
. Then the original length a is cut into two parts, say x and a - x, and then
Rollout 40, Act: -2.376
\n\nSecond term: -10 ��^n/(1 - ��) + 10/(
Rollout 41, Act: -2.385
count separately for each case. That is, count all 1-digit numbers, 2-digit numbers,
Rollout 56, Act: -2.396
a right-angled triangle at A0, with angle at A1 being 60°, which
Rollout 41, Act: -2.418
other digit is '0'.\n\nSo for each 'a' (from 1-9), we can
Rollout 5, Act: -2.430
125(x + 175)]/(x + 75) = 12
Rollout 41, Act: -2.442
count the number of three-digit numbers that use both a and b, arranged in some order. Each position
Rollout 70, Act: -2.487
as many digits as possible, with the highest possible digits.\n\nAlternatively, perhaps it's better to find N
Rollout 16, Act: -2.520
we can use the formula that for the maximum of independent random variables,\n\nE[T] = sum_{k
Rollout 33, Act: -2.570
=2, the first two positions must not both be from {1,2}. Wait, because if
Rollout 91, Act: -2.587
C.\n\nTo find angle at B: compute the angle between BA and BC.\n\nSimilarly, the angle at
Rollout 16, Act: -2.676
can be tricky. The expectation of the maximum of independent geometrically distributed variables isn't straightforward. In the
Rollout 5, Act: -2.947
+ 175)(125H)/(75 + x) = 125

DOWN_PROJ

Top 16 Positive Activations
Rollout 91, Act: 3.465
(t -1 )*0 - (-t )*(-2 ) | = |0 -2t |=
Rollout 45, Act: 3.295
1)(-2) + b*(-1)(-1) + c*(-2) +4\n\n
Rollout 91, Act: 3.272
(t -1 )*0 - (-t )*(-2 ) |= |-2t | =2t
Rollout 68, Act: 3.010
4= 4-8 +2k +4= 2k. So f(-2)=
Rollout 45, Act: 2.938
1)(-2) +8*(-1)(-1) +0*(-2) +4=
Rollout 81, Act: 2.918
3)(17) + (–h)(–h)= -51 + h2\n\n|
Rollout 55, Act: 2.863
would be 0*0 + (-1)*(-1) + 0*0 = 1
Rollout 90, Act: 2.844
digits, M is one digit, and L is four digits. For example, N=100
Rollout 81, Act: 2.821
(-3 - 0) = -h / (-3) = h/3.\n\nSimilarly, the
Rollout 96, Act: 2.812
69 =27 - (-12)*(-526)/169\n\nCompute x
Rollout 81, Act: 2.808
17)+ (-51/11)(-51/11)= -51 +
Rollout 91, Act: 2.796
t +1)(2) + (t)(0) = 2(t +1)\n\n|BA
Rollout 90, Act: 2.786
9, so:\n\nN �� (-M * (-112)) mod 1009
Rollout 17, Act: 2.735
6≈ Calculate:\n\n400*3.1416=1256.6
Rollout 91, Act: 2.723
|(t -1)(0) - t(-2)| = |0 + 2t|
Rollout 91, Act: 2.699
{BC},$ then find the value of $|\cot B - \cot C|.$\n<|im_start|>
Top 16 Negative Activations
Rollout 3, Act: -2.204
25,921;162^2=26,244. So
Rollout 47, Act: -2.205
2)\n\nWe already found v_air = (4/3) vt. Let's substitute:\n\np0
Rollout 89, Act: -2.207
they flipped the ratio. Maybe girls are 3/5. Let's imagine. If girls are3
Rollout 89, Act: -2.215
3 of the number of boys? Because 3/5 = boys, so per unit is 1
Rollout 84, Act: -2.229
drunk. Hence, that the wasted is 2/9 of the original wasted amount in volume.\n\nGiven
Rollout 34, Act: -2.230
x(t) - 2 ��_{0}^{t} e^{-2(t - s)}
Rollout 73, Act: -2.231
S has 11 elements, indices 0 to 10, so in our sum, j
Rollout 32, Act: -2.251
atrix}\n\]\n\nNow, compute M2:\n\n\[\n\mathbf{M}^2 =
Rollout 89, Act: -2.276
100, then girls would be (2/3)*100 = 66.
Rollout 95, Act: -2.298
their areas are (k/2450)^2 * Area ABC. Then, the area of
Rollout 8, Act: -2.301
5−r} or G^{5−r} R^r. So two options.\n\nIn terms
Rollout 7, Act: -2.303
3P=2N => P= (2/3)N.\n\nYes that's correct. So
Rollout 90, Act: -2.343
018.\n\n2018*100=201,800.
Rollout 34, Act: -2.371
constants: 4 -1 -3 = 0\n\nSo,\n\n= e^{-2t} [
Rollout 73, Act: -2.375
1 to n, so subtract the k=0 term (which is 0), so it's the
Rollout 24, Act: -2.432
allow a smaller set. Hence, they must be incomparable.\n\nTherefore, there's a possibility to build

Layer 36

GATE_PROJ

Top 16 Positive Activations
Rollout 25, Act: 3.454
is known.\n\nGiven that, proceed step by step.\n\nFirst, step 1: solve for r from
Rollout 53, Act: 3.423
)\n\nLet me write these with variables for easier calculation.\n\nCompute the area as:\n\nFirst, compute sum of
Rollout 96, Act: 3.392
697 ...(2)\n\nSolve this system.\n\nIt might be tedious but let me write them directly
Rollout 14, Act: 3.366
Therefore, determining λ and μ to satisfy the constraints.\n\nSo suppose x ranges up to x_max=1
Rollout 33, Act: 3.319
.\n\nSo I need to compute each of these terms. Let's start analyzing each |A_k|.\n\n
Rollout 56, Act: 3.286
.\n\nTherefore, this gives a recursion for each extension.\n\nGiven that, perhaps the series of areas can be
Rollout 96, Act: 3.254
1740 / 169.\n\nWait, let me compute that. Divide numerator and
Rollout 52, Act: 3.244
smaller integers.\n\nSo let's proceed step by step. Compute for each k, and check if n is
Rollout 53, Act: 3.237
. We can use these equations to solve for them.\n\nFirstly, subtract equation 3 from equation
Rollout 73, Act: 3.236
2^j.\n\nLet me evaluate each term separately.\n\nFirst, sum_{j=1}^{1
Rollout 96, Act: 3.234
seems rather messy. Let me check my computations again.\n\nEquation1:5x -12y
Rollout 96, Act: 3.230
need to solve these two equations for t and s.\n\nMultiply equation(1) by12:1
Rollout 20, Act: 3.227
*43*47.\n\nCalculating that:\n\nMultiply step by step:\n\n2*3=6
Rollout 99, Act: 3.226
So, let's calculate these parameters step by step.\n\nGiven our data points:\n\nx0 = 0
Rollout 28, Act: 3.222
(3)^2*(3)^2=0 ?\n\nWait no, with a=3, b=
Rollout 50, Act: 3.205
the coefficients of \(t\) and the constant terms.\n\nFirst, let's compute \((\log_{
Top 16 Negative Activations
Rollout 6, Act: -0.904
}$ . Show that there exists a sparse set $S_L$ and a deterministic polynomial time TM
Rollout 48, Act: -0.909
19E (since 57,344 is 0xE000,
Rollout 39, Act: -0.914
integer \[N = 9 + 99 + 999 + 999
Rollout 45, Act: -0.920
3 + m x^2 +2m x y +m y^2 +n x +n
Rollout 34, Act: -0.920
[\ny(t) = -4 + e^{-t} + Ce^{t}\n\]\n\nUsing the
Rollout 57, Act: -0.924
open. Therefore, since U is open, its complement C is closed. For the closed set C,
Rollout 69, Act: -0.952
2 sin2 theta = 1 - cos 2 theta. Therefore:\n\n1 - 4 sin2
Rollout 18, Act: -0.975
Z2 → Z/xZ defined by φ(k, m) = m - k y mod x.
Rollout 93, Act: -1.003
hev polynomials’ extrema are at cos(kπ/n) for k=0,1,...,
Rollout 63, Act: -1.007
, 1) = f(1, 3) = 3 \)\n- \( f(
Rollout 74, Act: -1.018
K_b might be the coefficient for the fore-and-aft component, and K_s for the athwart
Rollout 26, Act: -1.040
:\n\nHg(l) + 1⁄2 O2(g) → HgO(s)\n\nBut
Rollout 31, Act: -1.056
real numbers $x, y$. For any pair of real numbers $a_{0}, a_{1
Rollout 93, Act: -1.075
om{2014}{1007} + \binom{2014
Rollout 74, Act: -1.102
by external magnetic fields, like those from a ship's own magnetism or nearby ferrous materials. These
Rollout 6, Act: -1.196
's theorem equivalence between \( \textbf{P}/\text{poly} \) and sparse or

UP_PROJ

Top 16 Positive Activations
Rollout 72, Act: 2.376
boxed, should be \boxed{2k} for integer k ≥1, but since n must be
Rollout 95, Act: 2.367
Triangle \(ABC\) has side lengths \(AB = 9\), \(BC = 5\sqrt
Rollout 20, Act: 2.342
is a multiple of the order of p modulo m for every prime p ≠ m.\n\nBut since that must
Rollout 89, Act: 2.313
that exactly \( \frac{2}{5} \) of the children in a class are girls and
Rollout 37, Act: 2.313
product of primes up to 2015 and their multiplicities. But perhaps for each prime p
Rollout 99, Act: 2.302
log_{10}(x) - x + 3 = 0 \) with initial points \(
Rollout 86, Act: 2.275
each function. However, actually, the selection is with repetition, so all permutations of these counts would be
Rollout 58, Act: 2.264
<|im_start|>answer\nGiven an exam with \( k > 1 \) parts, where each part \( i
Rollout 19, Act: 2.261
k \) if 2 and \( k \) are coprime. But 2 and \(
Rollout 64, Act: 2.247
12=2009+3, so 2^{2012} -1
Rollout 60, Act: 2.217
source emits sound of frequency \( f_0 = 492 \, \text{Hz}
Rollout 99, Act: 2.204
calculate the differences and coefficients for the quadratic polynomial:\n\[\n\begin{align*}\nh_0
Rollout 45, Act: 2.197
\nGiven the function \( f: \mathbb{Z} \to \mathbb{Z}
Rollout 65, Act: 2.191
\leq c \) where \( b = 1995 \), we need to determine
Rollout 97, Act: 2.156
. The rate constant \(k\) is \(5 \times 10^{-4} \, \
Rollout 64, Act: 2.149
\) defined by the recurrence relation \(a_n = 3a_{n-1} - 2
Top 16 Negative Activations
Rollout 36, Act: -2.205
0)\n- D: (0, 0, s)\n- G: Since AG is a
Rollout 36, Act: -2.213
(s, 0, 0)\n\nC(0, s, 0)\n\nD(0,
Rollout 9, Act: -2.215
5), (10,0), (10,10). The overlapping area between this triangle
Rollout 81, Act: -2.230
A at (0,h) and D at (0,0), then BD is 3 and DC
Rollout 36, Act: -2.238
+ z^2 = 18000\n\n√[x^2 + y^2
Rollout 56, Act: -2.246
0 to A3 [...] and the new pasture is triangle A1A2A3. Next, she
Rollout 36, Act: -2.300
(0, s, 0)\n\nD(0, 0, s)\n\nG(s, s
Rollout 9, Act: -2.304
(10,0)\nM is (10,10)\nG is (10 -
Rollout 9, Act: -2.307
0,10), and E at (10,0). That's consistent.\n\nNow triangle G
Rollout 38, Act: -2.327
0), southwest (L, W), southeast (0, W). That might be complicated because x usually
Rollout 38, Act: -2.398
(L,W), and southeast corner is at (0,W). Wait, hold on: Since the longer
Rollout 36, Act: -2.407
s, 0)\n\nD(0, 0, s)\n\nG(s, s, s)\n\n
Rollout 9, Act: -2.445
E=(10,0), M=(10,10), G=(10 - h
Rollout 36, Act: -2.482
- 2sy + s2 = 18000\n\nEquation (3): x
Rollout 36, Act: -2.531
2 + z2 = 18000 => (x2 + y2 + z2
Rollout 36, Act: -2.537
sy + s2 = 18000\n\nEquation (3): x2 + y

DOWN_PROJ

Top 16 Positive Activations
Rollout 45, Act: 2.732
) = -1 +2 ⇒k =1\n\nTherefore, k=1.\n\nThen from Equation4
Rollout 75, Act: 2.601
,102,42,22,12,32. This is eight squares
Rollout 9, Act: 2.599
= 100 => y = 5.\n\nTherefore, the y-coordinate of point G must be
Rollout 62, Act: 2.565
another number is semiprime (two primes) or triprime (three), then the overall factor
Rollout 93, Act: 2.516
03.5. So (503.5 + 0.5)/100
Rollout 38, Act: 2.504
find the ratio of the field's length to its width to the distance from D to the southeast corner.
Rollout 75, Act: 2.437
,102,42,22,12. Which is eight squares, need to
Rollout 14, Act: 2.422
int_0^\infty f(x)^2 dx$,\n\n$K = \int_0^\
Rollout 75, Act: 2.422
2,42,22,12,32. This is eight squares. If we
Rollout 75, Act: 2.416
+102 +42 +22 +12 +32. That's nine squares
Rollout 95, Act: 2.400
/3 �� 8.333...\n\nThen x = 25/3. Then
Rollout 95, Act: 2.382
0/18 = 25/3 �� 8.333...\n\nThen x
Rollout 23, Act: 2.370
70=10^(-30.413)= around 10^-30
Rollout 53, Act: 2.360
.5 /6 =18.25\n\nTherefore, a= sqrt(73/4
Rollout 8, Act: 2.357
probability that Kathy will be happy is $\frac{m}{n}$ , where $m$ and $
Rollout 14, Act: 2.324
1})^{1/(2a -1)}}\n\nFrom K_λ=1: λ^{a
Top 16 Negative Activations
Rollout 28, Act: -1.984
-Y2)\n\nThird term:4*(X2 +2XY +Y2)/2=2
Rollout 20, Act: -1.987
1681 mod43:43*39=1677, so
Rollout 6, Act: -1.998
allows L to be reduced. So probably the intended answer is to cite/to use this theorem.\n\nHence
Rollout 65, Act: -2.001
Set up ratios x/a = b/b = c/c ⇒x/a =1⇒x=a, which
Rollout 19, Act: -2.028
and83. Hence, only divisors ≥6 are83 itself. Therefore, k=8
Rollout 1, Act: -2.031
circular.\n\nAlternatively, setting t = ||x|| squared, you get equations:\n\n-⟨x, y
Rollout 45, Act: -2.039
1)x +4m for all x ≥0 integers.\n\nRewritten as x3 +4x2
Rollout 8, Act: -2.047
actually, for k=2, is the number 1200 per sequence? Wait, no
Rollout 46, Act: -2.084
, when they say "in terms of U," they accept expressing it as the gradient of U adjusted by
Rollout 6, Act: -2.111
, but here the advice is fixed for all inputs of length n, and the machine runs in poly-time
Rollout 3, Act: -2.173
2)^2 = 170^2 - 4*170 + 4 =
Rollout 71, Act: -2.195
ac)x - abc= x3 -0x2 + (-2011)x + (-abc
Rollout 1, Act: -2.203
||2. Therefore:\n\nd2 = (d2)/2 + (d2)/2 -
Rollout 83, Act: -2.241
, and any rotated copy differs.\n\nIn standard Burnside approach, when considering labelings under rotation with distinct
Rollout 22, Act: -2.414
)^2 = 1,7002 - 2*1,700*6
Rollout 12, Act: -2.613
if that helps. Alternatively, substitution s = t2 - ν2?\n\nAlternatively, substitution φ where t

Layer 37

GATE_PROJ

Top 16 Positive Activations
Rollout 50, Act: 3.950
01) = 11.\n\nWhich checks out. Therefore, the value of \(x\) is
Rollout 5, Act: 3.781
5). Is that right?\n\nWait, let me check the calculations again step by step. Probably, I
Rollout 51, Act: 3.765
-\theta)|\ dx. \]\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this integral
Rollout 41, Act: 3.738
+ 576\n\nLet's compute that step by step:\n\nFirst, 9 +90
Rollout 16, Act: 3.724
.032 = 6.193548...\n\nk=6: C(
Rollout 17, Act: 3.712
I}|$ (in SI units).\n<|im_start|>assistant\n<|im_start|>think\nAlright, let me try to tackle
Rollout 67, Act: 3.704
1020=1848. Correct.\n\nGood. So c_n is modeled by this
Rollout 67, Act: 3.670
-420=1020. Correct. So equation 3 checks out given equations
Rollout 13, Act: 3.601
a dollar value to the nearest cent.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Joanie took a
Rollout 92, Act: 3.598
(-0.615 m). Let's compute this. First, convert centimeters to meters.\n\n
Rollout 59, Act: 3.593
. Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so the problem is about
Rollout 99, Act: 3.585
the decimal value of x_3?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work
Rollout 13, Act: 3.582
.12^4. Let me do this step by step.\n1.12^1 =
Rollout 52, Act: 3.561
50=2014. Yep, correct. So valid. So the sequence from -1
Rollout 40, Act: 3.549
19.1421356.\n\nSo the greatest integer less than or equal to
Rollout 20, Act: 3.542
licative, the general case would follow. Let me verify that.\n\nSuppose that for every prime power p
Top 16 Negative Activations
Rollout 66, Act: -2.511
2 = c2, so c2 - a2 = b2, thus (c - a)(
Rollout 37, Act: -2.524
can model the determinant as the product of the differences multiplied by a symmetric function. Then use the fact that
Rollout 77, Act: -2.527
) = \frac{52}{52 - 14} \cdot f(14
Rollout 66, Act: -2.532
that the sides were a= m2 -n2, b=2mn, but when in reality
Rollout 66, Act: -2.538
with opposite parity, a= m2 -n2, b= 2mn is given as a
Rollout 38, Act: -2.614
0,0). The length is L (west direction, so negative x-axis) and width W (
Rollout 12, Act: -2.621
ν}^1 [ (t2 - ν2)^{-1/2} (1 - t
Rollout 66, Act: -2.684
its sides are of the form \(m^2 - n^2\), \(2mn\),
Rollout 12, Act: -2.693
expression: E(k) - (1 -k2) K(k). If this is approaching1 -
Rollout 77, Act: -2.708
express this in terms of f(a, b - a)? Because if I can do that, maybe I
Rollout 38, Act: -2.724
origin (0,0). Let x-axis be west (so negative direction is west), y-axis be
Rollout 77, Act: -2.762
we can replace it with (a, b - a). Repeating until we get to (a,
Rollout 12, Act: -2.831
ν}^1 ( (t2 - ν2)^{-1/2} (1 - t
Rollout 66, Act: -2.985
\[\n \frac{ab}{c - a} = 6p\n \]\n
Rollout 66, Act: -3.179
of the form \(m^2 - n^2\), \(2mn\), and \(m
Rollout 66, Act: -3.189
primitive triples via the formula (m2 -n2, 2mn, m2 +n2

UP_PROJ

Top 16 Positive Activations
Rollout 5, Act: 3.582
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nOne base of a trapez
Rollout 34, Act: 3.582
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nFind continuous functions $x(t
Rollout 47, Act: 3.582
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA train with cross-sectional area $
Rollout 43, Act: 3.582
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nThere is a C-NOT gate
Rollout 78, Act: 3.582
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nJon and Steve ride their bicycles along
Rollout 63, Act: 3.582
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $P$ denote
Rollout 60, Act: 3.582
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA stationary source emits sound of frequency
Rollout 94, Act: 3.582
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nDetermine which set of states mentioned
Rollout 84, Act: 3.582
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nGary purchased a large beverage, but
Rollout 55, Act: 3.582
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet\n\[\mathbf{
Rollout 32, Act: 3.575
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $\omega$ be a non
Rollout 40, Act: 3.575
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA particle is located on the coordinate
Rollout 35, Act: 3.575
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $P(x)$ be a
Rollout 58, Act: 3.575
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nPatchouli is taking an exam
Rollout 50, Act: 3.575
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nThere is a positive real number $
Rollout 59, Act: 3.575
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nJackie and Phil have two fair
Top 16 Negative Activations
Rollout 72, Act: -2.504
- 1 = 0 \), so the sum is zero.\n9. Similarly, combinations with two
Rollout 71, Act: -2.505
Sum of squares is ��a2= (Σa)^2 - 2Σab=
Rollout 27, Act: -2.514
1/2)m v_z2, then ⟨E⟩ = (1/2)m ⟨v_z
Rollout 11, Act: -2.516
p is equal to the sum of the variables plus the sum of their products two at a time. That
Rollout 89, Act: -2.559
\( \frac{3}{5} \) of the children are boys, which we know is
Rollout 95, Act: -2.576
similar to ABC, with ratio x, then their areas are x2 * Area ABC. But in the
Rollout 11, Act: -2.587
equal to the sum of the variables plus the sum of their products two at a time. That form makes
Rollout 85, Act: -2.587
. Therefore, in both cases, there are two numbers of one parity and one of the other. So
Rollout 9, Act: -2.589
be equal. Therefore, the triangle has two equal sides (GE and GM) with base EM.\n\nSo
Rollout 76, Act: -2.590
square (2,2).\n\nTherefore, the intersection of all three is the single square (2,2
Rollout 95, Act: -2.624
similar to triangle ABC. Therefore, the ratio of areas would be proportional to the square of scaling factor.\n\n
Rollout 86, Act: -2.630
x dx= -pi/4. Therefore, -∫x sin2x dx = -(-pi
Rollout 89, Act: -2.689
= T divided by 5, girls=2 units, boys=3 units. Therefore, per unit
Rollout 89, Act: -2.730
33.333... so 2/5 equals 66.666...\n\n
Rollout 71, Act: -2.816
such that they sum to zero, their pairwise products sum to -2011, and their product
Rollout 71, Act: -2.854
abc= -m.\n\nSum of squares is ��a2= (Σa)^2 -

DOWN_PROJ

Top 16 Positive Activations
Rollout 18, Act: 2.312
sides by -7:\n\n5b +4c = -3\n\nSo, 5b +4
Rollout 12, Act: 2.117
+ [ nu2 K(k) ] / ( (1 - nu2) nu )\n\nSimplify terms
Rollout 12, Act: 2.098
) + [ nu2 K(k) ] / ( (1 - nu2) nu )\n\nSimplify
Rollout 2, Act: 2.067
8.\n\nTherefore, equation becomes:\n\n(5m +5)≡5 mod8\n\nSubtract
Rollout 18, Act: 2.048
, n) in Z2 such that pm + qn = 0 for some integers p and q
Rollout 6, Act: 2.046
textbf{NP} \).\n\n**Answer:** \n\( \textbf{P} = \text
Rollout 11, Act: 2.032
^{iθ}, c=1/(r2 e^{i2θ}). So a = b =
Rollout 71, Act: 2.020
8044 are say 89^2=7921, 90
Rollout 17, Act: 2.007
3/(A·s2)) / m3 ) = kg/(A·s2) = T
Rollout 62, Act: 1.999
=(4)/2=2 or (-2)/2=-1≡12 mod13. So
Rollout 89, Act: 1.975
, maybe 2/5th is 2 fifths—Wait, units.\n\nWait, maybe it
Rollout 45, Act: 1.973
Therefore,\n\n2a + b + 3c = 14 [Equation 2]\n\nWait
Rollout 53, Act: 1.968
1.5. Multiply both sides by 3:\n\nsqrt( c^2 + d^2 )
Rollout 58, Act: 1.962
i +1 )) * (i/(i +1 ))^{i -1} / ( (i
Rollout 17, Act: 1.960
Ω·s, but perhaps better in base units: J = kg·m2/s2, so
Rollout 98, Act: 1.956
K or M-type, right? So something's wrong here. A stars are hotter and more massive.
Top 16 Negative Activations
Rollout 43, Act: -2.403
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nThere is a C-NOT gate
Rollout 47, Act: -2.403
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA train with cross-sectional area $
Rollout 55, Act: -2.403
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet\n\[\mathbf{
Rollout 63, Act: -2.403
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nLet $P$ denote
Rollout 60, Act: -2.403
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nA stationary source emits sound of frequency
Rollout 78, Act: -2.403
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user\nJon and Steve ride their bicycles along
Rollout 55, Act: -2.405
because cos(π/6) = ��3/2 and sin(π/6) =
Rollout 30, Act: -2.406
[1/(a_k -1 ) -1/a_k ] = 1/(a_k(a_k
Rollout 30, Act: -2.416
n+1} -1/a_{n+3}). Then this was equal to [b_{n
Rollout 30, Act: -2.420
1/a_{n+1} - 1/a_{n+3}) as a difference s_n
Rollout 30, Act: -2.477
(1/a_{n+1} -1/a_{n+3}). Then this was equal to
Rollout 56, Act: -2.514
= |1*(√3 - (-√3/2)) + 0*(-√3/2
Rollout 66, Act: -2.514
2 +n2 ) - (m2 -n2 ) ) = gcd( m2 +n
Rollout 91, Act: -2.600
[ (t +1 ) - (1 - t ) ] / t = ( t +1 -
Rollout 95, Act: -2.701
12$ . Points $A = P_{0}, P_{1}, P_{2}, ...
Rollout 56, Act: -2.736
7/6*(√3 - (-√3/2 ))| / 2\n\nSimplify term

Layer 38

GATE_PROJ

Top 16 Positive Activations
Rollout 77, Act: 3.415
,52)=364.\n\n**Final Answer**\n\boxed{364}\n\n<|im_start|>answer
Rollout 36, Act: 3.381
: \boxed{192}\n\n**Final Answer**\n\boxed{192}\n\n<|im_start|>answer
Rollout 58, Act: 3.372
037171}.\n\n**Final Answer**\n\boxed{203717
Rollout 60, Act: 3.355
the answer should be 6 Hz.\n\n**Final Answer**\n\boxed{6}\n\n<|im_start|>answer\nA
Rollout 67, Act: 3.346
as concluded. Therefore, correct answer.\n\n**Final Answer**\n\boxed{348}\n\n<|im_start|>answer
Rollout 63, Act: 3.333
f(p,q) = pq \).\n\n**Final Answer**\n\boxed{f(p,q) = pq
Rollout 68, Act: 3.329
the desired answer is501.\n\n**Final Answer**\n\boxed{501}\n\n<|im_start|>answer
Rollout 38, Act: 3.324
is \boxed{61}。\n\n**Final Answer**\n\boxed{61}\n\n<|im_start|>answer\n
Rollout 62, Act: 3.299
\boxed{132}。\n\n**Final Answer**\n\boxed{132}\n\n<|im_start|>answer
Rollout 56, Act: 3.296
sqrt3. Therefore, we conclude.\n\n**Final Answer**\n\boxed{\sqrt{3}}\n\n<|im_start|>answer
Rollout 93, Act: 3.292
4}}} since the clearest expression.\n\n**Final Answer**\n\boxed{\dfrac{100
Rollout 59, Act: 3.289
the answer is 515.\n\n**Final Answer**\n\boxed{515}\n\n<|im_start|>answer
Rollout 66, Act: 3.287
correct answer is 182.\n\n**Final Answer**\n\boxed{182}\n\n<|im_start|>answer
Rollout 70, Act: 3.283
\boxed{315}.\n\n**Final Answer**\n\boxed{315}\n\n<|im_start|>answer
Rollout 16, Act: 3.281
53.\n\nTherefore, box that.\n\n**Final Answer**\n\boxed{789953
Rollout 71, Act: 3.276
is \boxed{98}.\n\n**Final Answer**\n\boxed{98}\n\n<|im_start|>answer\n
Top 16 Negative Activations
Rollout 31, Act: -0.824
1 -√5)^2]/4=(1 -2√5 +5)/4=(6 -
Rollout 49, Act: -0.824
a σ-compact subspace is a countable union of closed (compact) sets.\n\nBut how does
Rollout 40, Act: -0.831
+ ��2) = 4 - 2 = 2\n\nNumerator:\n\n[ (1
Rollout 87, Act: -0.842
.5): r=sqrt(0.25+0.25+0.25
Rollout 99, Act: -0.885
3362240124 -2.65731324
Rollout 17, Act: -0.899
=1; ��=π, u=-1. So integral becomes:\n\n∫_{-1}
Rollout 4, Act: -0.903
{eq}} \) is:\n\n\[\n\frac{\Delta C_{\text{eq}}}{
Rollout 89, Act: -0.907
500/3 ��200/3≈66.666...\n\nCannot
Rollout 40, Act: -0.935
)(2 + ��2) = 4 - 2 = 2\n\nNumerator:\n\n[
Rollout 70, Act: -0.958
0.3567\n\nSo, (-1.2528)/(0.35
Rollout 49, Act: -1.004
of X. Then X’’ is the set of limit points of X’, and so on. If
Rollout 31, Act: -1.006
2*(1 -√5)/2=(1 -5)/4= (-4)/4= -
Rollout 42, Act: -1.040
to form H2O+, which leaves to form a tertiary carbocation at the central carbon. Now,
Rollout 12, Act: -1.059
f(\nu)= \int_1^{\frac{1}{\nu}} \frac{dx
Rollout 75, Act: -1.079
the resulting 2k= (S -3)/4=2*(a2 +b2 +
Rollout 71, Act: -1.087
12d2 -12d ��1 -0≡1 mod4. Therefore, k

UP_PROJ

Top 16 Positive Activations
Rollout 1, Act: 2.398
)/2 = (3d2 / 2)/2 = 3d2/4.\n\n〈
Rollout 83, Act: 2.306
! for each group's permutations) => 3! * (3! * 3! *
Rollout 11, Act: 2.282
). Thus, p= (t1 +1/t1 ) + (t2 +1/t2
Rollout 38, Act: 2.268
is x/2⇒x=2*(x/2)=x. So no, it's L
Rollout 17, Act: 2.252
/A * J/T = (T·m/A) * (J/T) = (m/A)*(
Rollout 11, Act: 2.233
= \overline{a} + \overline{b} + \overline{c}
Rollout 28, Act: 2.232
= -B k / (2h -a) ... (1a)\n\nFrom equation (2):\n\n
Rollout 49, Act: 2.220
2^{|X| * \aleph_0} = 2^{|X|}. But
Rollout 12, Act: 2.216
Compute 1 - (t2 - ν2)/(1 - ν2) )\n\n= (1 -
Rollout 11, Act: 2.197
. Let's write c as 1/(ab). Then q becomes a/b + b/(1/(
Rollout 77, Act: 2.188
3+5)*15=8*15=120\n\nRight-hand side:5
Rollout 11, Act: 2.183
ab). So\n\n\overline{a b} = \frac{1}{a b} →
Rollout 69, Act: 2.181
i}, 3 theta_i =30° /3^{i}= theta_{i -1}\
Rollout 11, Act: 2.180
a*b ) =1/(a^2 z). Hence, c =1/(a^2 z
Rollout 11, Act: 2.165
hold if abc = 1/(a b c ), implying (abc)^2 = 1, so
Rollout 85, Act: 2.165
(2,0)=4. So set {2,3,4} ✓.\n\nCheck (x
Top 16 Negative Activations
Rollout 5, Act: -2.169
181.25\n\]\nThe greatest integer not exceeding 181.25
Rollout 50, Act: -2.174
\[\n\boxed{112}\n\]<|im_end|>
Rollout 61, Act: -2.177
, the fraction is in simplest form. Therefore, \( p = 133 \) and \(
Rollout 32, Act: -2.186
+ \mathbf{M}^2 + \mathbf{M}^3 + \dots
Rollout 41, Act: -2.187
927\)\n\nThe number of positive integers less than 10,000 that have
Rollout 45, Act: -2.201
\boxed{-1}. However, the question is the greatest possible value of m, so (b)
Rollout 73, Act: -2.203
398. Hence, the required remainder is 398. Therefore, the answer is
Rollout 95, Act: -2.208
(\overline{P_kQ_k}\) with rational lengths.\n\n1. **Coordinates and Similar Tri
Rollout 21, Act: -2.216
\) and \( S \subseteq X \).\n\n### Final Answers\n\n(a) \boxed{S \
Rollout 22, Act: -2.218
m/s} \)\n\n**Final Answer**:\n\[\n\boxed{1694} \
Rollout 13, Act: -2.219
12\n\]\n\n\[\n\boxed{187.12}\n\]<|im_end|>
Rollout 9, Act: -2.230
area common to triangle \(GEM\) and square \(AIME\) is 80 square units.
Rollout 41, Act: -2.232
927.\n\n**Final Answer**\nThe number of positive integers less than 10,00
Rollout 41, Act: -2.254
than 10,000 that have at most two different digits is \(\boxed{9
Rollout 44, Act: -2.296
positive. Statement [D] is incorrect.\n\nFinal Answer: \(\boxed{ABC}\)<|im_end|>
Rollout 82, Act: -2.309
{2}\n\]\n\nThe codewords are:\n- \( \frac{16}{69

DOWN_PROJ

Top 16 Positive Activations
Rollout 91, Act: 2.523
(t -1 )*0 - (-t )*(-2 ) | = |0 -2t |=
Rollout 28, Act: 2.498
/(2k -b)) *b ) /a )\n\nLet me write B on both sides (assuming
Rollout 91, Act: 2.449
(t -1 )*0 - (-t )*(-2 ) |= |-2t | =2t
Rollout 31, Act: 2.372
0\n\nSince this must hold for all possible a0 and a1, which means the coefficients of a
Rollout 81, Act: 2.320
AB: m1= (0 -h)/(-17 -0)=h/17.\n\n
Rollout 83, Act: 2.318
the second gives (r_{i+1} + r_{i+2} + r_{i
Rollout 64, Act: 2.231
+2*( (a2 -a1)/2 ) =2a1 -a2 +a
Rollout 7, Act: 2.220
5-point increase, every participant's score is increased by 5. Therefore, the new total score
Rollout 56, Act: 2.205
Δy= (11√3)/sqrt(559) * sqrt(559
Rollout 1, Act: 2.203
/ 2)/2 = 3d2/4.\n\n〈y, y〉 = ||y
Rollout 30, Act: 2.193
b_n ] / [a_{n+1}(a_{n+1} -1 ) ]
Rollout 75, Act: 2.187
3 mod 4. Alternatively, if we have seven squares (if we needed seven), but here it
Rollout 67, Act: 2.180
.\n\nLet me denote the general term c_n as An2 + Bn + C. Then given that
Rollout 96, Act: 2.177
69 =27 - (-12)*(-526)/169\n\nCompute x
Rollout 96, Act: 2.141
697*895:\n\nCompute 700*895=626
Rollout 46, Act: 2.140
0) and M2 is at position (-b, 0) in the rotating frame. But
Top 16 Negative Activations
Rollout 10, Act: -2.098
of numbers n_i where each n_i is a losing position, and then show if we can construct n
Rollout 28, Act: -2.103
Remember, a hyperbola has \( B^2 - 4AC > 0 \).\n\nFrom
Rollout 10, Act: -2.113
,1,2,3,...\n\nTherefore, the losing positions are numbers of the form 5k and
Rollout 49, Act: -2.119
continuum. The original problem states that X must be countable.\n\nSo to bridge the gap between 2
Rollout 82, Act: -2.121
8s.\n\nWait, more precisely:\n\nWait, initial probabilities:12,10,8
Rollout 66, Act: -2.176
. Since in a right-angled triangle, the altitude projects the hypotenuse into segments proportional to the
Rollout 93, Act: -2.186
ime, then yes, but k ranges from 1 to 1007, and as
Rollout 10, Act: -2.199
uent to 0 or 2 modulo5 are losing positions. Now, how to check if this pattern
Rollout 66, Act: -2.203
Given that in a right-angled triangle, the altitude to the hypotenuse is h = (leg
Rollout 10, Act: -2.219
7≡2 mod5\n\nSo actually, the losing positions are numbers congruent to 0 or
Rollout 69, Act: -2.246
)/cosθ cos3θ ] = sin2 theta / co[s3 theta,\n\nSo alpha (cos
Rollout 38, Act: -2.257
r2 -625r +168=0.\n\nUsing quadratic formula:\n\nr=(6
Rollout 93, Act: -2.282
2 + C(2014,1007}= if there's a identity as
Rollout 38, Act: -2.382
2 -1250 r +336 =0\n\nLet’s solve for r with quadratic
Rollout 66, Act: -2.876
recall that in a right-angled triangle, the altitude to the hypotenuse has a length equal to
Rollout 62, Act: -4.277
m ism=132, which 11^3*13. Which has four

Layer 39

GATE_PROJ

Top 16 Positive Activations
Rollout 46, Act: 1.251
-\frac{G M_{2}}{\rho_{2}}-\frac{\omega^{2}}{
Rollout 21, Act: 1.230
which is a finite nonempty string formed from a set $x_1, x_2
Rollout 40, Act: 1.212
coordinates, |p| + |q|.\n\nHmm, okay. Let's break down what each move
Rollout 60, Act: 1.141
's about the Doppler effect and beat frequency. Hmm, let's start by understanding the question.\n\nSo
Rollout 3, Act: 1.125
calculate the mean molecular speed of radon gas. Hmm, first, I remember there's a formula for
Rollout 99, Act: 1.106
need to find the decimal value of x3. Hmm, okay. \n\nFirst, I remember that the
Rollout 91, Act: 1.076
the value of |cot B - cot C|. Hmm, right. Okay, first, let me visualize
Rollout 52, Act: 1.062
. We need to find the smallest yummy integer. Hmm, interesting. So I need to figure out what
Rollout 48, Act: 1.054
00 008B 5BC6 AEE0 . Does the header contains error?\n
Rollout 60, Act: 1.043
. The sound reflects off the car and comes back to the source, where it's superimposed with
Rollout 42, Act: 1.040
produced when cyclobutyl(cyclopropyl)methanol reacts with phosphoric acid in water
Rollout 93, Act: 1.037
of (πk/1007). Hmm. Let me write that down to make it clearer
Rollout 87, Act: 1.035
density rho, and each side is length a. Hmm. Let me start by recalling some basic concepts about
Rollout 48, Act: 1.028
0 008B 5BC6 AEE0 . Does the header contains error?\n<|im_start|>
Rollout 97, Act: 1.021
mathrm{O}_{2}(g)$ is started in a closed cylinder under isothermal isochoric condition
Rollout 72, Act: 1.013
\( n \) whose sum is 1. Hmm, roots of unity...so these are complex numbers
Top 16 Negative Activations
Rollout 22, Act: -3.418
0). \n\nSince 1,6942 = 2,869,63
Rollout 73, Act: -3.421
is indeed 398. Let's compute 16398 - 16*
Rollout 3, Act: -3.430
K (unless stated otherwise), perhaps answer around \boxed{169} m/s. But need
Rollout 71, Act: -3.449
Is 1417 a perfect square? 372=1369,
Rollout 16, Act: -3.452
096 divided by 217. 217*5=1085
Rollout 3, Act: -3.475
50)=161.4, as 161^2=25,9
Rollout 96, Act: -3.510
8 ��2197=3 since 2197*3=659
Rollout 81, Act: -3.513
4,721=0.\n\nUse quadratic formula:\n\nx = [17,242
Rollout 89, Act: -3.548
/3 is acceptable.\n\nTherefore written boxed as \boxed{\dfrac{200}{3}}
Rollout 3, Act: -3.597
m/s. Thus, present answer boxed as \boxed{169 \text{ m/s}}
Rollout 22, Act: -3.641
). \n\nSince 1,6942 = 2,869,636
Rollout 26, Act: -3.653
, so two decimal places, the answer is \boxed{90.39}\n\n\n**Final Answer
Rollout 90, Act: -3.657
8: Let's compute 2018×50,000=100
Rollout 60, Act: -3.689
divided by 82 is 6, since 82*6 = 492,
Rollout 3, Act: -3.829
the mean molecular speed of radon gas is \boxed{170} m/s if we round
Rollout 3, Act: -3.965
this question is ambiguous. Nevertheless, likely answer \boxed{170} m/s. Alternatively,

UP_PROJ

Top 16 Positive Activations
Rollout 59, Act: 2.771
n = 123 + 392 = 515\).\n\n\[\n\
Rollout 39, Act: 2.755
N = \frac{10^{322} - 10}{9} -
Rollout 41, Act: 2.751
9 + 90 + 252 + 576 = 927\
Rollout 68, Act: 2.741
\): Integers from \(-500\) to \(-1\) give 500
Rollout 40, Act: 2.732
5i + 5(1 + i(\sqrt{2} + 1))(1 + i
Rollout 84, Act: 2.725
)\n\]\n\nExpanding and simplifying:\n\n\[\n9 - 36x = 4 -
Rollout 8, Act: 2.722
\(240 + 7200 = 7440\).\n\nThe probability is
Rollout 61, Act: 2.717
+ 13 + 7 + 1 = 133 \).\n\nThus, the probability
Rollout 58, Act: 2.711
).\n - Simplifying, each term for part \( i \) reduces to \( P(0)
Rollout 32, Act: 2.685
669 \cdot 6 + 4 = 4014 + 4 =
Rollout 41, Act: 2.679
9 + 216 + 27 = 252\)\n\n4. **4
Rollout 8, Act: 2.674
2400 + 1200 = 7200\).\n\nAdding both cases
Rollout 61, Act: 2.671
1 \)\n\nSumming these, we get \( 37 + 31 + 2
Rollout 64, Act: 2.671
}^8 \left(2^{2012} - k(2^{200
Rollout 41, Act: 2.653
\)\n\nTotal: \(9 + 90 + 252 + 576 =
Rollout 20, Act: 2.653
First, factorize 2021: \( 2021 = 43 \
Top 16 Negative Activations
Rollout 84, Act: -1.994
2). Wait, but that's different.\n\nWait, depending on how you read it: "twice
Rollout 96, Act: -1.997
system, point A is on the positive x-axis. Wait, actually, in the new system, since
Rollout 16, Act: -1.999
4 days. That seems low. How?\n\nWait, but in our problem, it's a bit different
Rollout 8, Act: -2.005
, examples include RRRRR and GGGGG.\n\nSo actually, if all red cards in the layout
Rollout 23, Act: -2.010
we are supposed to use the "partial half-life" concept where the decay constants are considered per decay mode
Rollout 56, Act: -2.012
then that triangle is replacing or being added? Wait, the problem says "the new pasture is triangle A
Rollout 97, Act: -2.021
amount in moles. It just gives a pressure. But since it's closed and under constant volume and
Rollout 89, Act: -2.045
5 of the number of boys. Hmm. Wait, let's check again.\n\nThe problem says: "
Rollout 56, Act: -2.061
A3, such that A3A0 is half of A2A0. So the original A
Rollout 47, Act: -2.097
tunnel walls are moving. Hmm, perhaps boundary conditions complicate things a bit here. Maybe we can assume
Rollout 47, Act: -2.116
the air that's moving with respect to the tunnel? Hmm.\n\nThis might be getting confusing. Let's
Rollout 47, Act: -2.118
moving towards the train at speed vt? Or is it the air that's moving with respect to the tunnel
Rollout 83, Act: -2.173
16; 1296? Wait, wait. Wait.\n\nWait.\n\nWait our mistake is
Rollout 96, Act: -2.229
I found are not the actual coordinates in the new system. Maybe something's wrong in the parametrization
Rollout 84, Act: -2.364
drank? Or twice the original purchased amount? Hmm, the problem says "twice as much". When
Rollout 77, Act: -2.436
, 4) should be 4? Wait, in my previous example, I computed f(2

DOWN_PROJ

Top 16 Positive Activations
Rollout 41, Act: 2.314
3,4), numbers of length n using exactly two different digits. For n=1, obviously zero
Rollout 37, Act: 2.283
entries in a row are the same). But a rank 1 matrix has determinant zero, because all rows
Rollout 1, Act: 2.255
d / sqrt(2) times sqrt(2) from center y, which is exactly d. Wait
Rollout 37, Act: 2.217
in a row are the same). But a rank 1 matrix has determinant zero, because all rows are
Rollout 20, Act: 2.179
similar steps as before: multiplying by (p-1), bringing p^{kn +1} -1
Rollout 17, Act: 2.176
m2/s2, so (kg·m2/s2 *m)/A = kg·m
Rollout 16, Act: 2.175
of n geometric(p) variables is equal to sum_{k=1}^{infty} [1
Rollout 37, Act: 2.170
are the same). But a rank 1 matrix has determinant zero, because all rows are multiples of each
Rollout 49, Act: 2.163
izable space is Lindelöf and metrizable, hence second countable). But 2nd
Rollout 15, Act: 2.161
0.\n\nThis is equivalent to x_{t+1}[i] = (x_{t}[
Rollout 97, Act: 2.143
t, integrate to in(-2k)Pt}\n\nThus Pressure is P= P0 e^{-2
Rollout 97, Act: 2.117
. The rate constant \(k\) is \(5 \times 10^{-4} \, \
Rollout 20, Act: 2.104
m. Then multiplying both sides by (p -1): p^{kn +1} - 1
Rollout 29, Act: 2.102
container open to the atmosphere, the gauge pressure at the bottom would just be rho*g*h, with h
Rollout 27, Act: 2.090
rms velocity:\n \[\n v_{z, \text{rms}} = \sqrt{\frac
Rollout 78, Act: 2.090
going east at 20 mph, Steve is going west at 20 mph. There are two
Top 16 Negative Activations
Rollout 17, Act: -2.126
Wait, 1.41*8=11.28. So 8π *
Rollout 19, Act: -2.145
but >=6? 83 itself is prime, so only 1 and83. So yes
Rollout 87, Act: -2.150
between π/4 and π/2, r goes from 0 to 1/sinθ.
Rollout 22, Act: -2.181
But here's another problem. Since the problem is in hydrogen gas, which in reality the molecules are di
Rollout 95, Act: -2.194
at k-th division along AB from A to B, divided into 2450 equal parts.
Rollout 22, Act: -2.204
2,800 �� 5,910,800.\n\nDifference:
Rollout 26, Act: -2.206
0.39 kJ/mol, and the absolute value |X| would be 90.
Rollout 3, Act: -2.227
/337≈0.047. Hence, 168.5 +0
Rollout 40, Act: -2.247
2 = i√2 + i*(2)/2 = i√2 + i1\n\nTherefore,
Rollout 26, Act: -2.253
) * R * T\n\nBut R is given as 8.3 J/K/mol, which is
Rollout 83, Act: -2.307
’s original residue\n\nGroup1’s residue = Group2’s original residue\n\nGroup2’s residue = Group
Rollout 83, Act: -2.310
to Group0. Similarly, rotation by 6 positions would cycle the groups in the reverse order.\n\nMeanwhile
Rollout 16, Act: -2.334
953-1904=49.\n\ngcd(68,49).\n\n6
Rollout 69, Act: -2.372
, i=2: 30° /9, that's 10/3°,
Rollout 83, Act: -2.503
Group0. Similarly, rotation by 6 positions would cycle the groups in the reverse order.\n\nMeanwhile,
Rollout 66, Act: -2.579
Pythagorean triple. The sides are thus of the form m2 - n2, 2mn

Layer 40

GATE_PROJ

Top 16 Positive Activations
Rollout 53, Act: 3.467
:\n\n1971 ��146:\n\nLet me factor numerator and denominator:\n\n197
Rollout 13, Act: 3.289
351936.\n\nBreaking it down:\n\n6000*1 = 60
Rollout 58, Act: 3.284
8 * 2019:\n\nFirst, 2018 * 2000
Rollout 53, Act: 3.264
1971 ��146:\n\n146*13=189
Rollout 96, Act: 3.264
06*169= Let's compute:\n\n4106*100=4
Rollout 22, Act: 3.257
6e23 kg/molecule\n\nCalculate that:\n\n0.002016 /
Rollout 96, Act: 3.252
check 1740 ��13:\n\n13*133 = 17
Rollout 13, Act: 3.249
36\n\nCalculating that value:\n\nFirst, 1 * 6000 = 6
Rollout 20, Act: 3.245
*43*47.\n\nCalculating that:\n\nMultiply step by step:\n\n2*3=6
Rollout 53, Act: 3.223
1971 divided by 146:\n\n1971 ��146:\n\n
Rollout 13, Act: 3.212
51936\n\nCalculating that value:\n\nFirst, 1 * 6000
Rollout 31, Act: 3.210
(x +1). Let's do polynomial division:\n\nx^3 -0x^2 -2
Rollout 4, Act: 3.210
,150 by 5025:\n\nFirst, 5025 * 1
Rollout 23, Act: 3.209
10 *945≈ Let's see: 945 *0.2=1
Rollout 61, Act: 3.208
1. Let's sum that. Let's compute:\n\n(37 +31) =68
Rollout 60, Act: 3.196
95 * 492\n\nFirst, 492 * 1 = 49
Top 16 Negative Activations
Rollout 87, Act: -1.054
:\n\n\[\nV = \frac{\rho a^2}{4\pi\epsilon_0}
Rollout 96, Act: -1.065
theta\) is the angle line \(L\) makes with the original x-axis. The rotation matrix for \
Rollout 44, Act: -1.070
is expanded reversibly with $\mathrm{T}_{1}=\mathrm{T}_{2}$, and (ii
Rollout 46, Act: -1.071
U\) with respect to \(x\) and \(y\) yield the gravitational and centrifugal forces:\n
Rollout 75, Act: -1.076
different positive odd integers can be represented as the sum of the squares of six (not necessarily different)
Rollout 99, Act: -1.082
h_1 &= x_2 - x_1 = 0.5, \\\n\delta_
Rollout 80, Act: -1.126
core by \( 1^\circ \mathrm{C} \):**\n \[\n \text
Rollout 69, Act: -1.135
(a)sin(b) = [cos(a - b) - cos(a + b)]/2.
Rollout 75, Act: -1.139
integers can be represented as the sum of the squares of six (not necessarily different) positive integers.\n
Rollout 99, Act: -1.156
- \delta_0}{h_1 + h_0} \approx -0.80
Rollout 50, Act: -1.197
of base formula, we express both sides in terms of base 10:\n\n\[\ny = \
Rollout 1, Act: -1.254
a Hilbert space such that the images are orthonormal.\n\nBut here, we're dealing within the
Rollout 44, Act: -1.275
. For an ideal gas, internal energy depends only on temperature, so if no work is done and no
Rollout 75, Act: -1.304
any three pairwise different positive odd integers can be represented as the sum of the squares of six (not necessarily
Rollout 75, Act: -1.335
three pairwise different positive odd integers can be represented as the sum of the squares of six (not necessarily
Rollout 10, Act: -1.578
impartial game under the normal play condition is equivalent to a nimber. Also, all subtraction games have periodic

UP_PROJ

Top 16 Positive Activations
Rollout 95, Act: 2.665
1,...,2449 as per the problem statement:\n\npoints A = P0; P1
Rollout 94, Act: 2.650
is NOT entangled. So answer choices (A) and (D) include b, but (D
Rollout 54, Act: 2.646
5,2,1,3 and4. So arranged as 5,2,1,3
Rollout 54, Act: 2.643
4 (methanol)\n\nThis would match the options (A = 5, 2,
Rollout 54, Act: 2.639
) → Answer choice A or C. The options given in the problem's first part (A's order
Rollout 94, Act: 2.616
, d is not entangled. So (A) is wrong.\n\n(B) a,c → both a
Rollout 94, Act: 2.612
is a candidate.\n\n(C) c,d → c is entangled, d is not. So (C
Rollout 54, Act: 2.591
first part (A's order):\n\nAnswer option A says A =5,2,1,3 and
Rollout 94, Act: 2.566
NOT entangled. So answer choices (A) and (D) include b, but (D)
Rollout 94, Act: 2.565
. So answer choices (A) and (D) include b, but (D) says a,b
Rollout 38, Act: 2.543
than length L, which contradicts the problem's statement that the longer side runs due west (i.e
Rollout 94, Act: 2.528
So answer choices (A) and (D) include b, but (D) says a,b.
Rollout 94, Act: 2.515
(b) is NOT entangled. So answer choices (A) and (D) include b,
Rollout 76, Act: 2.511
1 and S2 are columns 1-3, but S1 is columns 1-2 and
Rollout 23, Act: 2.511
15 Bq, which is option (D) 117.555, close
Rollout 98, Act: 2.508
.\n\nThus, 3 valid systems. But answer options are (A)1, (B)3
Top 16 Negative Activations
Rollout 93, Act: -2.051
combinatorial expressions.\n\nLet me verify: Consider the sum S = ��_{k=1}
Rollout 93, Act: -2.068
when dealing with sums over roots of unity or trigonometric sums, sometimes symmetry or properties of complex numbers
Rollout 93, Act: -2.069
, exponents with cosine can sometimes be dealt with using power-reduction formulas. Since the exponent here is
Rollout 78, Act: -2.079
. L / (v_w + 20) = 1/60\n\nTherefore, from
Rollout 50, Act: -2.080
logarithmic identities that might help here. The change of base formula is useful for comparing logarithms with different
Rollout 1, Act: -2.080
2).\n\nSo, how do we show that such a y exists?\n\nIn finite dimensions, if I have
Rollout 1, Act: -2.086
y. So the main idea is to find a point y such that all x in S are at distance
Rollout 81, Act: -2.110
geometry, we can model this.\n\nAlternatively, maybe coordinate geometry can help here. Let me place point A
Rollout 91, Act: -2.131
.\n\nBut perhaps it's better to assign coordinates. Let me place the triangle in coordinate system. Let me
Rollout 38, Act: -2.149
, initially heading south, and Cao bikes in a straight line across the field to a point D on
Rollout 91, Act: -2.163
, if the median AD makes an angle of 45 degrees with side BC, then find the value
Rollout 58, Act: -2.166
) = 2018 * P(0). We need to solve for k.\n\nTo calculate
Rollout 50, Act: -2.181
complicated because of all these logarithmic terms. Maybe I can simplify each logarithm expression? Let's expand
Rollout 49, Act: -2.212
statement.\n\nAlright, so given X is Hausdorff and uncountable, I need to find
Rollout 72, Act: -2.239
Maybe it's better to convert this into equations.\n\nLet’s represent ��5 as \( e^{2
Rollout 93, Act: -2.275
, perhaps we can consider using the beta function or integral approximation. But given that the exponents are discrete

DOWN_PROJ

Top 16 Positive Activations
Rollout 95, Act: 2.588
P_{k-1}$ and $P_{k+1}$ for $k = 1,
Rollout 1, Act: 2.529
1 + r2 >= l and |r1 - r2| <= l.\n\nIn our case,
Rollout 53, Act: 2.509
proceed. Shoelace formula:\n\nSum over i (x_i y_{i+1} - x
Rollout 14, Act: 2.321
up a functional:\n\n$$\n\mathcal{L} = \left( \int f(x)
Rollout 14, Act: 2.248
\int f(x) dx \right)^3 - \lambda \left( \int f(x)^
Rollout 53, Act: 2.209
. Shoelace formula:\n\nSum over i (x_i y_{i+1} - x_{
Rollout 81, Act: 2.194
alpha + tan beta]/[1 - tan alpha tan beta]\n\nWe are given tan(angle BAC)=
Rollout 14, Act: 2.190
functional to maximize is ��f dx -λ ��f2 dx - μ ��x f
Rollout 53, Act: 2.171
Shoelace formula:\n\nSum over i (x_i y_{i+1} - x_{i
Rollout 16, Act: 2.164
infty (1/2)^j =2, which matches.\n\nSimilarly, for n=2:
Rollout 89, Act: 2.161
So from here, multiplying both sides by 5/3 to isolate T:\n\nT = 10
Rollout 34, Act: 2.156
to t.\n\nd/dt [x(t)] = d/dt [1] + d/dt
Rollout 37, Act: 2.149
2-1)(3-1)(3-2) = 1*2*1=2
Rollout 45, Act: 2.147
4 + b -5 = -9 + b, which matches the previous expression. So both agreements,
Rollout 14, Act: 2.126
, we can set up a functional:\n\n$$\n\mathcal{L} = \left( \
Rollout 63, Act: 2.114
4}f(2, 1) \) both sides:\n\n\[\n\frac{3}{
Top 16 Negative Activations
Rollout 79, Act: -2.063
see. If "noticed" is the anagram indicator, then "minor changes" has to be an
Rollout 48, Act: -2.071
19E.\n\nNow, sum is 0xE19E. Now, add the last word
Rollout 23, Act: -2.091
-32. Then multiply by 5e34:\n\nN(t)=5e34 *
Rollout 95, Act: -2.094
(-2,√671)/ (5√3 ).\n\n The unit vector in BC direction: (-
Rollout 22, Act: -2.099
1e6 to eliminate decimals:\n\n18,168.7768 *
Rollout 48, Act: -2.105
=57,758 → 0xE19E (since 57,3
Rollout 14, Act: -2.119
^{-1/3}| dx ≤ ||f||_3 ||x^{1/3}||_
Rollout 80, Act: -2.121
* 7.238 is 12.8 * 7 = 89.
Rollout 14, Act: -2.126
||f(x)x^{1/2} ||_2 * ||x^{-1/2}||
Rollout 14, Act: -2.137
Then Hölder gives:\n\nI ≤ ||f||_2 ||1||_4 ||1||_
Rollout 4, Act: -2.162
V separately.\n\nStarting with C1: 2000 ± 10 pF. The
Rollout 14, Act: -2.199
\cdot 1 \le \|f\|_a \|1\|_b,\n\nBut as
Rollout 14, Act: -2.225
||f(x)x^(-1/3)||_a *||x^(1/3)||_
Rollout 48, Act: -2.269
612\n\nSum:\n\nStarting from the right:\n\n0 + 2 = 2\n\n0 +
Rollout 46, Act: -2.364
that M1 is located at ( -μ, 0 ) and M2 is at ( 1
Rollout 70, Act: -2.825
of N in base-7 (which are <=6), so the leading digit of N (base-

Layer 41

GATE_PROJ

Top 16 Positive Activations
Rollout 22, Act: 1.558
298 K) gives about 1,769 m/s.\n\nGiven the problem says "
Rollout 27, Act: 1.503
, need another variable.\n\nAlternatively, critical temperature may be calculated via equating thermal energy k T_c to
Rollout 27, Act: 1.497
Wait. Alternatively maybe, because the given ΔE_rms is 5.54e-1
Rollout 22, Act: 1.427
8 K (25°C):≈1,769 m/s\n\nAt 300
Rollout 46, Act: 1.343
be an obstacle.\n\nAlternatively, the problem may be considered in an inertial frame where U does not include
Rollout 27, Act: 1.342
the kinetic energy per particle to compute temperature and labeling that as T_c. So proceeding under this assumption,
Rollout 46, Act: 1.317
, that would not be the case.\n\nAlternatively, maybe they expect recognition that in the co-rotating
Rollout 46, Act: 1.305
have an error, and actually we are to assume that the terms are correct as gradients without Coriolis
Rollout 46, Act: 1.288
additional relationships, perhaps simplifying the scenario.\n\nAlternatively, in some formulations, maybe U is taken to include
Rollout 3, Act: 1.286
8 K (25°C) yields ~168.5 m/s. Therefore, since the
Rollout 3, Act: 1.264
and rounded to three sig figs is 169 m/s. Thus, present answer boxed as
Rollout 3, Act: 1.245
compute", so it's expected to compute.\n\nAlternatively, perhaps check if there is an established average speed for
Rollout 97, Act: 1.242
t=ln(10)/k=4605.\n\nThis is because in the first case
Rollout 89, Act: 1.237
teacher may suspect answer is 66 or 67, depends on their approach.\n\nBut given that
Rollout 27, Act: 1.237
. Therefore perhaps the problem is oversimplified, assuming that with V_Be calculated, T_c is when
Rollout 3, Act: 1.235
precision needed.\n\nAlternatively, maybe the original question assumes a specific temperature but forgot to mention. If the user
Top 16 Negative Activations
Rollout 20, Act: -3.629
81=1,948, 1800*1000=
Rollout 96, Act: -3.633
9*15=2535. 2630 -2535=
Rollout 60, Act: -3.657
by 328:\n\n328 goes into 332 once, remainder 4.
Rollout 90, Act: -3.673
0=201,800. 224,000 -20
Rollout 23, Act: -3.673
5 *0.2=189. 945*0.031≈
Rollout 20, Act: -3.688
(46*20=920, 920 +46=96
Rollout 19, Act: -3.692
3081 mod83.\n\nCompute 83*37=3071
Rollout 60, Act: -3.708
492\n\n492 * 0.01 = 4.92\n\n
Rollout 17, Act: -3.727
138.190, and 0.1077*3.14
Rollout 95, Act: -3.768
3).\n\nCompute area: (1/2)| (0*(0 - ��671/
Rollout 17, Act: -3.769
≈138.190, and 0.1077*3.1
Rollout 96, Act: -3.782
*3=85,683; 85,683 <88,
Rollout 13, Act: -3.783
0.57 * 6000 = ?\n\nWait, maybe better to process 60
Rollout 39, Act: -3.907
788=7092, because 9*700=6300
Rollout 17, Act: -4.065
41: 8*1 =8, 8*0.41=3.2
Rollout 13, Act: -4.092
3000 → 0.57 * 6000 = ?\n\nWait, maybe

UP_PROJ

Top 16 Positive Activations
Rollout 65, Act: 2.480
k*a, k*1995, k*c in any order.\n\nSo, suppose after cutting
Rollout 68, Act: 2.470
. And since their product positive, both roots are positive.\n\nTherefore, in k>8, quadratic equation
Rollout 0, Act: 2.438
but their exponents are 1, which is odd. Actually, wait nvm, the exponents
Rollout 62, Act: 2.417
? Because 11* p*q would be three primes, but not four. Unless factor with exponent
Rollout 39, Act: 2.403
990/9 = 110, which is 11 followed by a zero
Rollout 96, Act: 2.400
direction of line L is (12/13,5/13). The unit vector perpendicular
Rollout 21, Act: 2.378
x -f(x)= -0.5 -0=-0.5. Ceil(-0.
Rollout 20, Act: 2.364
}) �� 1*1*...1 �� 1 mod 2021.
Rollout 9, Act: 2.356
). The altitude is h, the horizontal distance from G to EM. So h is the length we need
Rollout 49, Act: 2.341
or set is compact, so it's σ-compact. Therefore, being a perfect set in our X
Rollout 56, Act: 2.331
1 and ��3, hypotenuse 2, angles 90°, 60
Rollout 57, Act: 2.322
are also open, hence all open sets are also closed. Thus, you would have that the topology is
Rollout 62, Act: 2.317
: 11*p*q*r, which is four primes. So need the (11k2
Rollout 65, Act: 2.316
the smaller prism must also be ka : kb : kc for some k < 1. Therefore, that
Rollout 49, Act: 2.312
of a compact metrizable space is sigma-compact? NO! Wait, example:\n\nIn the Cant
Rollout 56, Act: 2.308
- Hypotenuse A1A2 = 2.\n\nSo coordinates:\n\n- Let’s set A0
Top 16 Negative Activations
Rollout 57, Act: -2.566
don't manage to find the separation for regular, perhaps I need to try a different approach. Let me
Rollout 24, Act: -2.594
much consideration without progress, I suspect the key insight requires constructing a injective mapping between rows of two minimal
Rollout 10, Act: -2.599
exist. As a contest problem, the solution may be found in realizing the periodicity failed, but that
Rollout 14, Act: -2.600
who set the problem must have an particular method in mind. Let's go back.\n\nAnother strategy. Use
Rollout 6, Act: -2.606
in circles here. Given the time constraints, maybe for problem 3, since it's known in complexity
Rollout 14, Act: -2.610
the person who set the problem must have an particular method in mind. Let's go back.\n\nAnother strategy
Rollout 15, Act: -2.613
, power configurations may make difference.\n\nAlternatively, perhaps answer is that n must be a power of2
Rollout 15, Act: -2.632
the possible patterns and their evolutions. Perhaps a solution exists where nが is a power of two?
Rollout 14, Act: -2.634
Either way, this isn't helping. Perhaps I need a different approach.\n\nAlternative idea: Use Hölder
Rollout 37, Act: -2.639
uncertain. Moreover, if given the problem must answer then official solution uses said prime conditions, so possibly
Rollout 24, Act: -2.647
proof, and after much consideration without progress, I suspect the key insight requires constructing a injective mapping between
Rollout 6, Act: -2.675
time.\n\nSo perhaps the answer is not straightforward, hence why reference to the P/poly theorem. The
Rollout 57, Act: -2.693
manage to find the separation for regular, perhaps I need to try a different approach. Let me recall that
Rollout 37, Act: -2.738
key insights (or lack thereof), I think the answer hinges on the fact that determinant(A) vanishes
Rollout 24, Act: -2.857
for a proof, and after much consideration without progress, I suspect the key insight requires constructing a injective
Rollout 10, Act: -2.946
infinitely many positions exist. As a contest problem, the solution may be found in realizing the periodicity failed

DOWN_PROJ

Top 16 Positive Activations
Rollout 48, Act: 2.825
= 9+14=23. 23 in hex is 17, which
Rollout 39, Act: 2.818
092, because 9*700=6300,\n\n9*80
Rollout 74, Act: 2.642
ft direction is at ��, and the athwartships is at �� + 90 degrees
Rollout 96, Act: 2.606
3=3,211\n\nSo 3.11... Hmm. Anyway, use fractions
Rollout 32, Act: 2.495
x3 = 1 are 1, ��, and ��2, where �� and
Rollout 34, Act: 2.487
) = -4 exp(0) + exp(-t) + C exp(t)\n\nSo,\n\ny
Rollout 96, Act: 2.417
=10,894.\n\n13*800=10,400
Rollout 96, Act: 2.401
remaining 2561. 13*197=2561. Yes
Rollout 22, Act: 2.387
458.933\n\nNow, 2,111 * 69 �
Rollout 70, Act: 2.376
0. Then, 20 divided by7=2, remainder6. Then, digits are1
Rollout 25, Act: 2.360
previous case with:\n\nOriginal mu = GM_earth, leading to perigee 9898
Rollout 70, Act: 2.359
7=1 rem2, 2/7=0 rem2. So digits 3 1
Rollout 36, Act: 2.354
^4 -30,624s2 +140,541,6
Rollout 16, Act: 2.346
is converted on or before day t is 1 - (1/2)^{t+1}
Rollout 76, Act: 2.335
-95: 512 -90=422, 422 -
Rollout 11, Act: 2.318
1, 1/a = bc, 1/b = ac, 1/c = ab, as
Top 16 Negative Activations
Rollout 10, Act: -1.951
between 3 and 2? Let's see:\n\n0 to2: +2\n\n2 to5
Rollout 81, Act: -1.954
h/-3= h/3. So since run is negative direction, from A to B delta x
Rollout 77, Act: -1.965
* f(14,10)= (24/10)*70=24
Rollout 77, Act: -1.976
8. Which is LCM(14,24)=168. Indeed. Then f
Rollout 73, Act: -1.980
total sum is 2036. Correct.\n\nSo 18434-20
Rollout 27, Act: -1.990
given by:\n\nT_c = ( (n^(2/3) h^2) / (2
Rollout 91, Act: -1.995
) = [2(t +1)] / [2 sqrt( (t +1)^2 + t
Rollout 71, Act: -2.020
yield answers, this triplet is likely the only possible one. Hence, absolute sum |39| +
Rollout 78, Act: -2.035
going east at 20 mph, Steve is going west at 20 mph. There are two
Rollout 82, Act: -2.049
)/69= (16+30=46;46+36=8
Rollout 56, Act: -2.101
�3).\n\nDistance A4A2:\n\nx difference: 0 -7/6= -7
Rollout 4, Act: -2.159
1200. Yep, that's right. So, C_eq is 120
Rollout 33, Act: -2.195
120=120+48=168, 168+3
Rollout 81, Act: -2.287
0^2 = 4900\n\n4*11*561 = 4
Rollout 73, Act: -2.355
0:\n\n10 +18=28\n\n28 +32=60\n\n6
Rollout 73, Act: -2.512
8\n\n28 +32=60\n\n60 +56=116\n\n

Layer 42

GATE_PROJ

Top 16 Positive Activations
Rollout 37, Act: 1.339
positive integers (positive integer excludes $0$ ). Let $$ A=\begin{pmatrix
Rollout 86, Act: 1.313
. That means we can use each function more than once, right? The goal is to choose four functions
Rollout 84, Act: 1.295
it, and m and n are coprime positive integers. Then, if he had purchased half as
Rollout 51, Act: 1.282
be a constant number such that $0\leq \theta \leq \pi.$ \n
Rollout 50, Act: 1.275
[\log_{20x} (22x)=\log_{2x} (20
Rollout 72, Act: 1.258
n \) for which there exist three (not necessarily distinct) roots of unity of order \( n \
Rollout 86, Act: 1.246
allowed. That means we can use each function more than once, right? The goal is to choose four
Rollout 13, Act: 1.207
4 years. How much more money will she owe if the interest compounds quarterly than if the interest compounds
Rollout 86, Act: 1.200
,\ \sin x,\ \cos x$ repeatedly. Find the minimum value of $\int_{
Rollout 80, Act: 1.191
mathrm{~g} / \mathrm{cm}^{3}$, and a specific heat of $
Rollout 6, Act: 1.175
{N} \mapsto \mathbb{N}$ such that for each $n \
Rollout 65, Act: 1.168
P_{},$ and both of which have nonzero volume. Given that $b=199
Rollout 28, Act: 1.151
term:\n\nFirst term: (u2 + 2uv +v2)/4\n\nSecond term:
Rollout 85, Act: 1.123
) must form the set {n, n+1, n+2} for some nonnegative integer
Rollout 86, Act: 1.116
repetition allowed. That means we can use each function more than once, right? The goal is to choose
Rollout 86, Act: 1.093
{x, sin x, cos x}, with repetition allowed. That means we can use each function more
Top 16 Negative Activations
Rollout 4, Act: -2.774
9 J=15 nJ.\n\nSimilarly, compute E_max=0.5*120
Rollout 28, Act: -2.775
, k) satisfying:\n\nFrom the center equations:\n\n1. 2A h + B k + D
Rollout 8, Act: -2.775
0P5).\n\nLet me try this approach.\n\nFirst, how many color sequences are "happy"?\n\nCase
Rollout 96, Act: -2.782
β.\n\nSo perhaps solving the problem via projection.\n\nFirst, projection of P onto line L.\n\nLine L
Rollout 8, Act: -2.792
), or all reds or all greens.\n\nSo first, we can calculate the number of such happy sequences
Rollout 97, Act: -2.796
Let’s let the amount dissociated at time t be such that the partial pressure of N2O5
Rollout 48, Act: -2.814
Does the header contains error?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this IPv4 header
Rollout 48, Act: -2.822
the one's complement.\n\nAlternatively, here's a step-by-step process:\n\nStart with a 32
Rollout 33, Act: -2.827
there's no k=1...0 condition.\n\nForn=2:\n\nPermutations[1,2
Rollout 97, Act: -2.868
1 atm. At time t, let the extent of reaction be x (in terms of pressure).
Rollout 56, Act: -2.869
: (0, -√3/2)\n\nTo compute the area of triangle A1A2A
Rollout 86, Act: -2.887
.\n\nSimilarly, integral of x2 cos2x dx can be handled analogously.\n\nUsing cos2x
Rollout 73, Act: -2.908
}(10 -i)2^i]\n\nFirst term: sum_{i=0}^{1
Rollout 93, Act: -2.943
we try to see a pattern here.\n\nm sum:\n\n1:1, 1=2^0
Rollout 48, Act: -3.026
0\n7. 0A00 = 0x0A00\n8.
Rollout 20, Act: -3.169
47.\n\nCalculating that:\n\nMultiply step by step:\n\n2*3=6\n\n6*7

UP_PROJ

Top 16 Positive Activations
Rollout 43, Act: 2.958
C lists exactly this. So I think the answer is option C.\n\n**Final Answer**\n\boxed{
Rollout 8, Act: 2.939
31/126, so m=31, n=126.\n\nWait
Rollout 84, Act: 2.900
he purchased. So the answer must be 37.\n\nTherefore, confident that the answer is \boxed
Rollout 74, Act: 2.842
deviation is small, then the approximation δθ �� (H_b sinθ + H_s cosθ
Rollout 92, Act: 2.811
, the answer should be approximately +3.78 diopters, rounded to two decimal places,
Rollout 84, Act: 2.743
of it, and m and n are coprime positive integers. Then, if he had purchased half
Rollout 73, Act: 2.741
the remainder is 398. So answer is 398.\n\nWait, but just to
Rollout 19, Act: 2.738
integersn. So seems the answer is just78. But wait, let me check for smaller numbers
Rollout 84, Act: 2.731
37.\n\nTherefore, confident that the answer is \boxed{37}.\n\n**Final Answer
Rollout 40, Act: 2.720
1421356 is 19. Therefore, the answer is 19.\n\n
Rollout 5, Act: 2.710
the greatest integer not exceeding that is 181. Therefore, the answer is 181
Rollout 8, Act: 2.696
126, so m=31, n=126.\n\nWait, but let's
Rollout 5, Act: 2.694
181.25 is 181. Therefore, final answer is 181
Rollout 92, Act: 2.691
rounding to two decimal places, 3.78 D.\n\nAlternatively, the problem might be considering the
Rollout 5, Act: 2.689
the greatest integer not exceeding this is 181.\n\nWait, this seems straightforward now, but maybe
Rollout 78, Act: 2.683
as m/n where m and n are coprime, and then compute m + n.\n\nAlright,
Top 16 Negative Activations
Rollout 38, Act: -1.834
Solving the quadratic equation after squaring both sides.\n\n6. **Finding Ratios**:\n -
Rollout 95, Act: -1.840
3, 2/3).\n\n Wait, wrong. Perpendicular vector to (-2/3, �
Rollout 93, Act: -1.845
theta\) as \(\frac{e^{i\theta} + e^{-i\theta}}{
Rollout 13, Act: -1.846
thousandths place shifted by four decimal places for multiplier position):\n0.056275\n
Rollout 11, Act: -1.848
\) and analyze the symmetry and properties of these roots.\n\n3. **Symmetric and Real Conditions**:\n
Rollout 14, Act: -1.861
Another idea: For each x, use the AM-GM inequality.\n\nSuppose for each x, we
Rollout 65, Act: -1.879
leads to equations from maintaining similarity requires that, through substitution of proportionality.\n\nSo, in case cuttingc
Rollout 11, Act: -1.901
and cyclic in \(a, b, c\).\n\n2. **Transformation and Simplification**:\n
Rollout 33, Act: -1.915
from the total permutations and adjust using inclusion-exclusion.\n\n2. **Applying Inclusion-Exclusion
Rollout 75, Act: -1.916
can be either 0, 1, or 4 modulo 8.\n\n2. **Example Decom
Rollout 33, Act: -1.938
m=5)**: Intersection of all \(A_1, A_2, A_3,
Rollout 75, Act: -1.940
)/4=0mod4? No, no:\n\nWait, let’s calculate 2k.\n\nActually
Rollout 70, Act: -1.969
i=0}^k d_i \cdot 7^i\n \]\n - This
Rollout 11, Act: -1.993
roots.\n\n3. **Symmetric and Real Conditions**:\n - Since \(p\) and \(q
Rollout 50, Act: -2.047
the ratio of the equations derived from the logarithmic terms. We find that:\n\n\[\n22x
Rollout 85, Act: -2.134
determine the labels for adjacent points and observe any patterns.\n\n2. **Analyzing Initial Points**:\n

DOWN_PROJ

Top 16 Positive Activations
Rollout 60, Act: 2.434
)\n\nThus, substituting f' into f'':\n\nf'' = f0 * (v + v
Rollout 0, Act: 2.430
of such coprime pairs is \(2^k\), where \(k\) is the number of
Rollout 20, Act: 2.383
47 are coprime, the Chinese Remainder Theorem says that the congruence mod
Rollout 54, Act: 2.358
on O, while propionate's charge is delocalized, making it a weaker nucleophile.\n\nSo
Rollout 91, Act: 2.325
B and C using the dot product formula.\n\nDot product formula: cos(theta) = (v ��
Rollout 22, Act: 2.287
2,111 * 69 �� 145, so 6
Rollout 15, Act: 2.256
matrix is nilpotent if some power of it is the zero matrix.\n\nSo, for M being the
Rollout 49, Act: 2.237
any subset A of X, when given the subspace topology, is σ-compact.\n\nIn particular,
Rollout 20, Act: 2.236
47. Therefore, by the Chinese Remainder Theorem, solving σ(an) ��
Rollout 74, Act: 2.183
the ship is headed at an angle �� from magnetic north, then the ship's magnetic fields would be
Rollout 82, Act: 2.170
�� 0.2319\n\np2 = 15/69 ��
Rollout 59, Act: 2.167
and B heads, C tails: (1/2)(1/2)(3/7) =
Rollout 69, Act: 2.160
rationals. Therefore, the coefficients of the irrational terms must be zero, and the rational part must equal
Rollout 54, Act: 2.136
-stabilized anion; the charge is delocalized, which makes it a weaker nucleophile compared
Rollout 23, Act: 2.129
but if answer option (D) is 117.555 Bq, perhaps it
Rollout 20, Act: 2.129
*3*7.\n\nCheck if2^42≡1mod43 yes.\n\nBut check order
Top 16 Negative Activations
Rollout 52, Act: -2.075
+1)/2= (38 -105)/2= (-67)/2=
Rollout 31, Act: -2.077
linear combination of a_{0} and a_{1}?\n\nWait here's an angle to approach:
Rollout 13, Act: -2.083
55\nMultiply as numbers:\n\n 11255 (ignore decimals for now)\nx
Rollout 85, Act: -2.094
are 2000*2 +2024 =6024 and 2
Rollout 82, Act: -2.106
code is just the initial symbol, no binary digits. So len_binary = 0. Therefore, contribution
Rollout 97, Act: -2.107
2ξ) + 2ξ + �� = n0 + ��. So total pressure
Rollout 72, Act: -2.127
1 + 1 - 1 = 1\n- 1 - 1 - 1 =
Rollout 75, Act: -2.128
4+4+1+4+4+1=16+4+4+1+
Rollout 62, Act: -2.133
+22k +1 -11k -1 +11=121k2
Rollout 66, Act: -2.140
12n2 +n, then m and n would have a parity determined by n. If n
Rollout 48, Act: -2.171
\n4. 4000\n5. 4006\n6. 0
Rollout 34, Act: -2.200
e^{-t} +3 e^t + e^{-t} -4 = 3 e^
Rollout 43, Act: -2.212
0⟩ + β|1⟩ - γ|1⟩) �� |0⟩, which simpl
Rollout 94, Act: -2.234
0> + |01> - |10> - |11> )\n\nAgain, ignoring
Rollout 14, Act: -2.280
$. So $f(x) = a - b x$. \n\nBut wait, we are assuming that $
Rollout 79, Act: -2.317
H, A, N, G, E, S. So S is there once. Okay. So

Layer 43

GATE_PROJ

Top 16 Positive Activations
Rollout 22, Act: 2.570
{1694} m/s.\n\n<|im_start|>answer\nTo compute the mean molecular speed \( v \
Rollout 50, Act: 2.525
Answer**\n\boxed{112}\n\n<|im_start|>answer\nGiven the equation:\n\n\[\n\log_{
Rollout 59, Act: 2.510
Answer**\n\boxed{515}\n\n<|im_start|>answer\nJackie and Phil each flip three coins:
Rollout 9, Act: 2.507
Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides of length
Rollout 0, Act: 2.503
Answer**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and
Rollout 65, Act: 2.500
Final Answer**\n\boxed{40}\n\n<|im_start|>answer\nGiven a right rectangular prism \( P \)
Rollout 20, Act: 2.484
\(\boxed{125}\).\n\n<|im_start|>answer\nTo solve the problem, we need to find
Rollout 71, Act: 2.478
Final Answer**\n\boxed{98}\n\n<|im_start|>answer\nGiven the polynomial \(x^3 -
Rollout 77, Act: 2.473
Answer**\n\boxed{364}\n\n<|im_start|>answer\nGiven the function \( f \) defined on
Rollout 8, Act: 2.464
Answer**\n\boxed{157}\n\n<|im_start|>answer\nKathy has 5 red cards and
Rollout 34, Act: 2.427
e^{-t} - 4}\).\n\n<|im_start|>answer\nTo solve the given integral equations for continuous functions
Rollout 61, Act: 2.427
Answer**\n\boxed{913}\n\n<|im_start|>answer\nGiven the set \( S \) of integers
Rollout 91, Act: 2.422
**Final Answer**\n\boxed{2}\n\n<|im_start|>answer\nGiven triangle \(ABC\) with median \(AD
Rollout 70, Act: 2.413
Answer**\n\boxed{315}\n\n<|im_start|>answer\nA positive integer \( N \) is called
Rollout 55, Act: 2.412
}{2} \end{pmatrix}}\n\n<|im_start|>answer\nThe given matrix \(\mathbf{A
Rollout 62, Act: 2.400
Answer**\n\boxed{132}\n\n<|im_start|>answer\nTo find the least positive integer \( m \
Top 16 Negative Activations
Rollout 61, Act: -3.123
3 +7 +1= indeed let's compute:\n\n37 + 31 is 68
Rollout 71, Act: -3.123
. 892=7921, 902=8100.
Rollout 56, Act: -3.130
Step 0: ~0.866\n\nStep 1: ~1.299
Rollout 35, Act: -3.141
inant: 49 +4=53, non-integer roots.\n\nD=-1: b
Rollout 67, Act: -3.142
3= -192 +540=348. Perfect.\n\nThus so the values
Rollout 71, Act: -3.146
. 782=6084, 792=6241.
Rollout 48, Act: -3.157
12 is 46,610. 17,204 + 4
Rollout 56, Act: -3.180
Step 1: ~1.299\n\nStep 2: ~1.511
Rollout 71, Act: -3.184
6, 452=2025. 1984/4=4
Rollout 71, Act: -3.185
(136)=11.66, not integer.\n\nWait, perhaps look for e such
Rollout 30, Act: -3.189
000 +28,210=4,058,210,
Rollout 71, Act: -3.218
? 372=1369, 382=1444,
Rollout 72, Act: -3.239
1 -0.5=0.5. Hmm.\n\nAlternatively, consider the imaginary parts. Wait
Rollout 71, Act: -3.280
, 452=2025. So 2008 is not a square
Rollout 11, Act: -3.298
a*b*c =k*k*(1/k2) = 1. Also, none of them are
Rollout 71, Act: -3.348
. 442=1936, 452=2025.

UP_PROJ

Top 16 Positive Activations
Rollout 49, Act: 2.845
to recall that in a first-countable, separable space, every open cover has a countable sub
Rollout 62, Act: 2.657
) plus the factors in the other number.\n\nWait, if 11 is a prime factor, the
Rollout 33, Act: 2.646
position as 6! = 720, but actually the first five elements must be exactly
Rollout 56, Act: 2.642
use is opposite the right angle. Wait, no, hypotenuse is the side opposite the right angle
Rollout 82, Act: 2.631
≈3.105 per codeword? Wait, no: group3's codewords
Rollout 8, Act: 2.627
! ) / (5 - (5 -r))! ; wait, maybe it's better to think
Rollout 58, Act: 2.613
i/(i + 1))^{i} ]... Wait, maybe not.\n\nWait, the entire
Rollout 21, Act: 2.548
x, f(x) - n = 0 for some n ∈ Z. But again, equations need
Rollout 79, Act: 2.538
letters: "Incident" has 8 letters, but "Room" is 4. Let's
Rollout 43, Act: 2.536
did they mention that the target is the first qubit instead of the second? Let me check.\n\nThe
Rollout 82, Act: 2.531
* total probability of the group). Wait, no, the total expected number is sum_{i=1
Rollout 65, Act: 2.524
1, c., need proportions either:\n\n a/k a= B1 /k*1995
Rollout 66, Act: 2.522
, no, all three are pairwise coprime? Wait, no. Wait, hypotenuse must
Rollout 56, Act: 2.518
otenuse is opposite the right angle. Wait, no, hypotenuse is the side opposite the right
Rollout 21, Act: 2.516
(x) - floor(x) = c, a constant, then f(x +1) -f(x
Rollout 56, Act: 2.491
is the side opposite the right angle. Wait, no: in a right-angled triangle, the sides
Top 16 Negative Activations
Rollout 38, Act: -2.045
\n\nThus, Equation1.\n\nEquation2 remains:\n\n(W +x)/6.2= sqrt(x
Rollout 81, Act: -2.053
21)^2 –17,242*(2601/121)+
Rollout 45, Act: -2.062
3a]\n\nEquation4:4k +2m + (4m -16k
Rollout 67, Act: -2.063
Bn + C\).\n\nUsing the given terms:\n- For \(n=1\): \(1
Rollout 4, Act: -2.067
25.2004=603*25 +603*0.
Rollout 67, Act: -2.070
*8 +1020= -72*64 + 3936 +
Rollout 7, Act: -2.076
5*(2N/3) = (75*2)/3 *N=150
Rollout 45, Act: -2.101
Equation 3]\n\nFrom f(2):4k +2m +n=12 [
Rollout 67, Act: -2.119
+ d)(p + q) = 1716. ...(2)\n\nFor n=3
Rollout 96, Act: -2.123
16*148=2316*(100+40+8)=
Rollout 34, Act: -2.125
) = \int_{0}^{t} e^{-2(t - s)} [2x(s
Rollout 12, Act: -2.144
+ u)^2 - ν2 ) (1 - (ν + u)^2 ) )\n\nSimplify
Rollout 81, Act: -2.154
01*484=2601*400 +2601*
Rollout 85, Act: -2.186
is to figure out all possible labels for the point (2000, 2024
Rollout 67, Act: -2.190
p + 2q) = 1848. ...(3)\n\nSo now we have three
Rollout 67, Act: -2.307
\]\n\[\nc_8 = -72 \cdot 64 + 393

DOWN_PROJ

Top 16 Positive Activations
Rollout 96, Act: 2.568
) = (1740/169, -1134/169
Rollout 81, Act: 2.387
/17:\n\ntan(theta) = |(-h/17 - h/3)/(1 +
Rollout 78, Act: 2.328
. Therefore, their relative speed is 620/9 + 20 = 62
Rollout 22, Act: 2.308
v_mean = sqrt(18,168.7768 / 0.0
Rollout 26, Act: 2.293
= -3.\n\nΔH = -296 kJ + (-3) * (8.
Rollout 37, Act: 2.292
would be a1^{k1}a2^{k2} - a1^{k2}
Rollout 36, Act: 2.270
sy + s2) = 36000 - 18000\nS
Rollout 96, Act: 2.255
_y) = (1740/169, -1134/16
Rollout 96, Act: 2.240
69. So OA_x=4056/169 -1740/
Rollout 96, Act: 2.213
from O(1740/169, -1134/169
Rollout 36, Act: 2.190
(1'):\n\nS = 36000 + 2s x - s2\n\n
Rollout 64, Act: 2.174
be positive:\n\na1 = A + 2B >0. Similarly, a2 =A +
Rollout 36, Act: 2.169
00/s) x = (18000/s + 7200/s)
Rollout 67, Act: 2.151
But from equation 1: p =1440 / a.\n\nSimilarly, from equation 2
Rollout 79, Act: 2.143
8,4)\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this crossword clue. The clue
Rollout 5, Act: 2.137
5H\n2. (x + 175)(H - h) = 12
Top 16 Negative Activations
Rollout 48, Act: -6.184
so it's perhaps easier to sum everything in 32 bits, then add the carryover bits afterward
Rollout 82, Act: -6.185
symbols. Thus, all codewords have length 2. Then, the expected number of symbols would
Rollout 97, Act: -6.199
as written is 2N2O5 -> 2N2O4}+O2}
Rollout 38, Act: -6.200
(L - x) (east) ] / 8.6 = [2L - x +
Rollout 35, Act: -6.213
D=−10, where (n−17)(n−24)=−10
Rollout 19, Act: -6.236
.\n\n- Then squared gives 100≡17modk.\n\n- Thus k divides83
Rollout 89, Act: -6.255
frac{2}{5} \times \frac{500}{3} = \frac{
Rollout 90, Act: -6.313
0 mod 2018: N*10^4 + M*10^3
Rollout 24, Act: -6.352
.\n\nBut even if |R|=|C| doesn't necessarily give same row number between different pairs.\n\n
Rollout 38, Act: -6.391
meters to the point (- (L - x), 0), then south W meters to (- (L
Rollout 21, Act: -6.410
would contain all integer-valued periodic functions with period 1 and h(0)=0 due to f
Rollout 83, Act: -6.501
arrangements fixed except identity.\n\nSame reasoning for rotation by 6.\n\nFor rotation by1 position - it affects
Rollout 53, Act: -6.589
= 2.5 which links to AC = 5. So three pieces: CE = 4
Rollout 21, Act: -6.788
So f would have to be a function with period 1. Then, in combination with other equations,
Rollout 14, Act: -6.942
dx ) * sqrt(∫A^infty 1 dx )= sqrt(J2)* sqrt(in
Rollout 29, Act: -7.189
compartment exerts higher pressure in that compartment, but doesn't affect the open compartment's pressure. If they

Layer 44

GATE_PROJ

Top 16 Positive Activations
Rollout 99, Act: 2.272
.5 \), and \( x_2 = 1 \), we need to perform two iterations of
Rollout 19, Act: 2.248
^3 \) is divided by \( n + 5 \), the remainder is 17.
Rollout 30, Act: 2.108
- 1$ for $n \geq 2$\n- $a_1 = a_
Rollout 2, Act: 2.086
we consider the product modulo 8 and modulo 125 separately, then combine the results using the
Rollout 55, Act: 2.079
3}}{2} & 0 & -\frac{1}{2} \\ 0 &
Rollout 37, Act: 2.036
& a_{2015}^{k_2} & \cdots & a_{2
Rollout 59, Act: 2.036
probability that they get the same number of heads.\n\nFirst, we determine the probability distribution for the number of
Rollout 50, Act: 2.014
, and then compute \(m + n\).\n\nFirst, let \(y = \log_{20
Rollout 82, Act: 2.001
9}, \frac{8}{69} \right)\n \]\n\n2. **Grouping
Rollout 78, Act: 1.999
find the length of each train.\n\nKey steps:\n1. Let \( L \) be the length of
Rollout 32, Act: 1.996
{M}^{2009}\).\n\nFirst, we note that \(\omega\) is a
Rollout 36, Act: 1.996
\sqrt{10}\), \(CP = 60\sqrt{5}\), \(DP
Rollout 88, Act: 1.991
of weights (excluding biases) in this layer:\n\n1. Each filter has dimensions 5x5x
Rollout 95, Act: 1.989
{671}k}{3 \times 2450}\right)\).\n -
Rollout 95, Act: 1.979
= Q_0, Q_1, Q_2, \ldots, Q_{24
Rollout 29, Act: 1.958
P \) at the bottom of the jug.\n\n1. **Understanding Gauge Pressure**: Gauge pressure is the
Top 16 Negative Activations
Rollout 59, Act: -3.373
2 = 16*24.5? Wait, 16 *24.5
Rollout 70, Act: -3.380
So digits are1 (for 350?), wait no:\n\nWait 363 divided by
Rollout 97, Act: -3.381
85/0.0005= approx 4605.17 seconds.\n\n
Rollout 71, Act: -3.386
<= 8044/12 �� 670.333, so
Rollout 14, Act: -3.388
)^{1/2 } )^{.1 } + sqrt(J)/2.\n\n Wait, actual calculation
Rollout 69, Act: -3.404
(theta_0)/cos(theta_{-1}) ?\n\nBut theta_{-1} is undefined. H
Rollout 54, Act: -3.416
weaker, that means HS− is a stronger base. Wait, wait. No: wait, the lower
Rollout 69, Act: -3.421
cos(3 theta' /3^j) ?\n\nWait, perhaps confused here. Let's make substitution
Rollout 26, Act: -3.422
3.5 kJ (four sig figs?)\n\nThen, the adjustment for Hg(l) is
Rollout 30, Act: -3.428
/ (a_{n+1 } )2 )\n\nWait, no:\n\nWait, if a_{n
Rollout 8, Act: -3.438
/ (5 - (5 -r))! ; wait, maybe it's better to think:\n\nWait
Rollout 55, Act: -3.540
3\*2018pi/6. Wait maybe that?\n\nWait, the rotation matrix for
Rollout 69, Act: -3.543
1}\times 10 to 3 scaling.\n\nAlternatively, maybe consider telescoping through trigonometric
Rollout 83, Act: -3.564
=1296/s=128, which is 1296/9=
Rollout 70, Act: -3.565
9: N=331 +16=331? Wait, wait N=3
Rollout 75, Act: -3.749
5+27+12=54+. Wait, no.\n\nWait, honestly, seems like

UP_PROJ

Top 16 Positive Activations
Rollout 89, Act: 1.365
33=66.666.\n\nHmm, so 66.6 recurring.\n\nIn
Rollout 36, Act: 1.351
*1,856=18,560,000\n\n1,8
Rollout 64, Act: 1.342
1000\n\nn=3: -998 +3996=29
Rollout 81, Act: 1.294
3, delta y= 0 - h= -h. hence, slope= delta y/delta
Rollout 23, Act: 1.270
1.925e32 Bq.\n\nBut again not the answer options.\n\nAlternatively, given
Rollout 82, Act: 1.251
4)/69≈2.087.\n\n same result. Therefore higher than2.\n\nHence
Rollout 62, Act: 1.247
=2. Nope. check if prime Maybe.\n\nOK perhaps back to approach.\n\nTurns out, I
Rollout 92, Act: 1.228
61.5 cm from the lens. So previous calculation is correct.\n\nI think the answer is
Rollout 75, Act: 1.214
four: 4. Total sum 9. Hmm, no. Maybe splitting into numbers where?\n\nWait
Rollout 75, Act: 1.204
, unless 2m -1 is square.\n\nHmm, perhaps the identity is this:\n\nm2 =
Rollout 36, Act: 1.202
00*1,856=18,560,000\n\n1
Rollout 22, Act: 1.186
) in m/s\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to calculate the mean molecular
Rollout 93, Act: 1.172
also be interpreted as (probability amplitudes?) Hmm. Alternatively, maybe we can use the identity that
Rollout 3, Act: 1.168
) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the mean molecular speed
Rollout 77, Act: 1.159
properties: \[f(x, x) = x,\; f(x, y) = f(y
Rollout 55, Act: 1.159
\renewcommand{\arraystretch}{1.5} \begin{pmatrix} \frac{\
Top 16 Negative Activations
Rollout 31, Act: -2.313
0 => x y = -1\n\nSo, substitute y from the first equation into the second equation:\n\n
Rollout 51, Act: -2.322
x| dx\n\nAlternatively, if I let t = x - π for the second integral, substitute:\n\n
Rollout 83, Act: -2.331
44.\n\nAlternatively, to verify consider specific small case, like total acceptable arrangements without symmetry=12
Rollout 40, Act: -2.331
1421356.\n\nSo the greatest integer less than or equal to 19.
Rollout 38, Act: -2.338
2857. Let's divide numerator and denominator by 48: 2304
Rollout 34, Act: -2.343
the initial condition. From the original integral equation, when t = 0:\n\nx(0) =
Rollout 12, Act: -2.355
vanish or something.\n\nAlternatively, maybe we can substitute variables to absorb the dependence on ν?\n\nAlternatively, instead
Rollout 12, Act: -2.357
, maybe we can perform a substitution in the integral to make the integral transform into another form where the dependence
Rollout 0, Act: -2.361
Let me check. The primes less than or equal to 20 are: 2, 3
Rollout 5, Act: -2.379
= 181.25, greatest integer not exceeding 181.25 is
Rollout 11, Act: -2.401
+ c/a. So let's express q in terms of a, b, c. Since abc =
Rollout 30, Act: -2.446
} b_n -1.\n\nBecause when n is replaced by n+1 in the b_{n+
Rollout 12, Act: -2.510
the variable of integration. Let me consider switching the variable to t, but not sure.\n\nWait, let
Rollout 58, Act: -2.521
= 2018. Therefore, the number of parts is 2018. Therefore
Rollout 7, Act: -2.639
is 56. Then, due to an error, all scores are increased by 5. After
Rollout 51, Act: -2.642
can make substitution to simplify things. Maybe shifting the variable to eliminate ��? Let's let y =

DOWN_PROJ

Top 16 Positive Activations
Rollout 70, Act: 2.746
306. Wait that's correct. So why earlier when I tried to divide 153
Rollout 73, Act: 2.660
+3=23. Wait, no, why +3? Because sum_{j=1}
Rollout 6, Act: 2.629
if the advice is generated in polynomial time, then why not just incorporate that polynomial-time computation into the main
Rollout 15, Act: 2.576
from all on, ends after two steps. Then why did I earlier thought it cycles?\n\nBecause I made
Rollout 99, Act: 2.567
first iteration. The question is ambiguously worded.\n\nHowever, given that it specifically asks for "the
Rollout 62, Act: 2.566
847, so m=(1+34)/2=17.5~ decimal.
Rollout 97, Act: 2.554
Which gives Y ~2.3.\n\nBut then why was the previous answer approx 4.6?\n\n
Rollout 9, Act: 2.547
confirm the positions. If the square is labeled AIME, moving around the square, then maybe A is
Rollout 7, Act: 2.540
itself isn't changed. Hmm. This is ambiguous.\n\nBut according to the problem statement: "due to
Rollout 68, Act: 2.538
other is not. So two cases:\n\nCase 1: The quadratic equation has exactly one real solution which
Rollout 76, Act: 2.534
3, wait 417: 4+1+7=12, so divisible by
Rollout 48, Act: 2.533
we should express the numbers as:\n\n0x E 1 9 E\n\n+0
Rollout 62, Act: 2.521
so 2081-2080=1. Not divisible. By 7=
Rollout 57, Act: 2.509
But then C must be open in the density topology.\n\nThis leads to the conclusion that in the density topology
Rollout 95, Act: 2.500
in lies the problem. Which is the correct relationship?\n\nHmm. If triangle APkQk is similar
Rollout 1, Act: 2.499
find such a y by fitting it accordingly.\n\nSo why in the previous example with translated set S', it
Top 16 Negative Activations
Rollout 62, Act: -2.264
-5 +4=24≡3≡3≠0\n\nm=6:36 -
Rollout 62, Act: -2.268
-6 +4=34≡6≡6≠0\n\nNo solution. Therefore, modulus
Rollout 0, Act: -2.270
}{b}\) and calculate the product of the resulting numerator and denominator. We need to determine how many
Rollout 64, Act: -2.275
2009} +a1 -k ≤0\n\nBut this seems not helpful. Alternatively,
Rollout 49, Act: -2.290
-finite open cover of X has a countable subcover. Wait, but not sure.\n\nAlternatively,
Rollout 50, Act: -2.322
terms to the left-hand side:\n\n\[\nt(\log_{10}22 + \log
Rollout 1, Act: -2.337
2 - 2c + d2)/2 = 0 ⇒ 0 = 0.\n\nThus
Rollout 10, Act: -2.344
2,...\n\nDifference between 0 and 2: 2\n\n2 to 5: 3\n\n
Rollout 82, Act: -2.356
)/69 = (16+15=31; 31+24=
Rollout 58, Act: -2.358
one question correct in the entire exam versus getting none correct. That seems like we have to compute these two
Rollout 41, Act: -2.365
number like 123, which has three different digits, would not be counted. Got it.\n\n
Rollout 75, Act: -2.368
9 split into 4+4+1 gives three more squares, and the +1 is already covered
Rollout 59, Act: -2.381
1+16.\n\n9+100=109; 109+1
Rollout 20, Act: -2.390
3,47.\n\nSum →2+3=5; 5+7=12;
Rollout 41, Act: -2.484
and 2), but 123 has three digits. So we need to count all three-digit
Rollout 21, Act: -2.531
f(x)(f(x) - 1) = 0, which forces f(x) to be

Layer 45

GATE_PROJ

Top 16 Positive Activations
Rollout 40, Act: 3.606
6 is 19. Therefore, the answer is 19.\n\nBut let me verify my steps
Rollout 27, Act: 3.588
. Given this confusion, perhaps the initial step is correct where ΔE_rms is (1/2
Rollout 0, Act: 3.551
, corresponding to \boxed{128}? Or wait, wait prime factors of 20
Rollout 2, Act: 3.527
the remainder is109.\n\nTherefore, answer:109.\n\nBut let me recheck all
Rollout 4, Act: 3.517
approximately 0.88%.\n\nWait, maybe 0.88 is okay. Alternatively, let
Rollout 87, Act: 3.515
] }\n\nSince I recall this from internet sources. Or it might be expressed using different constants.\n\nAlternatively,
Rollout 99, Act: 3.508
714012\n\nSo x3≈2.15733466
Rollout 3, Act: 3.506
168 or 169. Maybe 168 m/s.\n\nBut checking further on
Rollout 64, Act: 3.487
2^{2011}) ? Or no, my prior calculations yielded that the total number is
Rollout 13, Act: 3.486
616\n\nTherefore, A_annual �� $9,441.12.
Rollout 38, Act: 3.471
24=61.\n\nTherefore, the answer is \boxed{61}。\n\n**Final Answer
Rollout 5, Act: 3.463
is 181. Therefore, the answer is 181. \n\nWait, but let
Rollout 60, Act: 3.460
6.00\n\nSo total f'' �� 492 + 6 = 4
Rollout 33, Act: 3.460
correct answer. Hence, I'd conclude the answer is\boxed{461}.\n\n**Final
Rollout 87, Act: 3.434
. Considering the options, I think the answer is likely:\n\nV = \boxed{ \dfrac{\
Rollout 5, Act: 3.411
is 181. Therefore, final answer is 181. \n\nI think that's
Top 16 Negative Activations
Rollout 39, Act: -1.167
\sum_{k=1}^{321} 1\n\]\n\nThe first sum is
Rollout 37, Act: -1.189
} & \cdots & a_2^{k_{2015}} \\\n\vdots
Rollout 37, Act: -1.195
2015}} \\\na_2^{k_1} & a_2^{k_
Rollout 41, Act: -1.197
10,000 have at most two different digits, we consider numbers of different lengths (1
Rollout 73, Act: -1.200
left( \sum_{i=0}^{j-1} 2^j - \sum
Rollout 73, Act: -1.216
0 \) to \( 2^{10} \). We need to consider all pairs \((
Rollout 53, Act: -1.221
((-a, 0)\) and \((a, 0)\), and point \(C\)
Rollout 39, Act: -1.275
} 10^k - \sum_{k=1}^{321} 1
Rollout 73, Act: -1.276
, \ldots, 2^{10}\} \), we need to find the sum of
Rollout 39, Act: -1.289
^{k-1} + 10^{k-2} + ... + 10^
Rollout 37, Act: -1.295
15}^{k_{2015}}\n\end{pmatrix}\n\]\n\n### Key
Rollout 99, Act: -1.314
x) - x + 3 = 0 \) with initial points \( x_0 =
Rollout 82, Act: -1.344
0, \( \frac{8}{69} \) is encoded as 81\n
Rollout 61, Act: -1.351
\( 2^k + 2^m \) (with \( k > m \)) is
Rollout 37, Act: -1.361
^{k_1} a_2^{k_2} ... a_2^{
Rollout 73, Act: -1.553
j - \sum_{i=0}^{j-1} 2^i \right)\n

UP_PROJ

Top 16 Positive Activations
Rollout 7, Act: 2.925
"due to an error in the wording of a question, all scores are increased by 5. At
Rollout 9, Act: 2.806
square. Then the triangle GEM overlaps the square, and their common area is 80 square units
Rollout 7, Act: 2.728
: These are participants with original score <65, average 56. Their total score is
Rollout 7, Act: 2.722
"due to an error in the wording of a question, all scores are increased by 5. At
Rollout 7, Act: 2.692
at 65. Wait, the scores are increased by 5, so original scores become original +
Rollout 15, Act: 2.674
rule doesn't mention the current state. The next state depends only on the number of neighbors. So regardless
Rollout 7, Act: 2.665
, 6 remaining. Original average of 2 students=62, so total score=62
Rollout 33, Act: 2.658
many such permutations are there? For the first three positions:\n\n- First two positions: 2! arrangements
Rollout 7, Act: 2.658
"due to an error in the wording of a question, all scores are increased by 5. At
Rollout 25, Act: 2.657
The problem states that when the asteroid's mass changes, the parameters v (speed at infinity) and r
Rollout 8, Act: 2.637
arranged in same color order, but different actual cards, it's still counts as the same color sequence.
Rollout 90, Act: 2.634
says that when you remove the middle digit from A, you get B. For A with an odd number
Rollout 58, Act: 2.626
multiply across questions.\n\nBut the problem is about the probability of getting exactly one question correct in the entire exam
Rollout 59, Act: 2.624
(X=k)]^2, so total probability is sum over k [P(X=k)]2.\n\nSo
Rollout 32, Act: 2.623
, and then summing all those totals together. So essentially, if S is the sum of matrices M
Rollout 89, Act: 2.621
the question is saying. It says that in a class, 2/5 of the children are girls
Top 16 Negative Activations
Rollout 79, Act: -1.682
7, not 8+4. Not matching.\n\nAlternatively, "Machine Organs". M, A
Rollout 49, Act: -1.689
able. Therefore, it is a counterexample.\n\nHence, X may still resist being second-countable
Rollout 79, Act: -1.720
isn't related. Letters don't fit. This isn't working. Perhaps I'm missing a common phrase
Rollout 49, Act: -1.722
But we proved X is first-countable, but don’t known if second-countable.\n\nBut X is
Rollout 79, Act: -1.724
but no. As before. What's another? Let's try "Harmonizing CES". Not that
Rollout 79, Act: -1.725
– but that's 10 letters. Hmm.\n\nAlternatively, perhaps "Minor changes" can be rearr
Rollout 79, Act: -1.729
we have only one E, no. So no.\n\nAnother guess: "Answering Machine". But that
Rollout 11, Act: -1.736
. No good.\n\nAlternatively, take a = \overline{1/b}, leading to modulus |a
Rollout 79, Act: -1.749
zine ��मे" – nope.\n\nWait, maybe "Hormones" — using
Rollout 21, Act: -1.755
, etc., but constants are like 1, 2, π, etc.\n\nWait, suppose we
Rollout 79, Act: -1.761
.\n\nAlternatively, "Agency Office" – but letters don't match.\n\nWait, perhaps "Inchoate
Rollout 79, Act: -1.770
12 letters. So Search Wing isn't correct.\n\nThen I hit upon the idea: maybe "Main
Rollout 49, Act: -1.790
. For example, in a locally compact Hausdorff σ-compact space, you can show it
Rollout 79, Act: -1.825
anagram of "minor changes": M, I, N, O, R, C, H,
Rollout 79, Act: -1.880
's see. Maybe "Chroning"? But we don't have a T. "Horming"?
Rollout 79, Act: -1.899
the answer is "Monitoring... something". But letters aren't matching. Alternatively, "Change Minor" –

DOWN_PROJ

Top 16 Positive Activations
Rollout 14, Act: 2.641
1/3} \right)^{2/ (3/2 -1)} } \right)^
Rollout 11, Act: 2.548
ate variables? Not sure. Maybe if |a| ≠ 1, but since abc = 1
Rollout 81, Act: 2.536
CAB is the angle at point A between points C, A, and B. Wait, maybe this coordinate
Rollout 78, Act: 2.505
1/13.5? Wait, no.\n\nWait, 440 divided by 9
Rollout 38, Act: 2.503
04/672=2304 �� 672= approx 3
Rollout 60, Act: 2.489
}$. The sound is reflected by a large car approaching the source with a speed of $2 \mathrm
Rollout 28, Act: 2.454
, the midpoint of OG? Or related to medians.\n\nWait, in the general coordinate system, G
Rollout 87, Act: 2.421
0 to a / cos �� or a / sin ��, depending on the angle, which
Rollout 32, Act: 2.391
2 +0 + (−ω2 from S) ?? Wait no. Let me recomputethe previous
Rollout 49, Act: 2.382
pace. But if X is uncountable and regular (since Hausdorff and hereditarily
Rollout 38, Act: 2.382
, but point D is not necessarily at the southwest corner. It's somewhere on the south edge, maybe
Rollout 61, Act: 2.379
different pairs (m1, k1) and (m2, k2), even if d=k
Rollout 54, Act: 2.362
. Wait, the pKa of cyclohexanol is around 19-20 (where
Rollout 71, Act: 2.352
105), (-3,-35), (-5,-21), (-7,-15
Rollout 26, Act: 2.350
problem, all data given seems to be precise to four significant figures. For example:\n\n- Heat capacity of
Rollout 8, Act: 2.339
9 * 8 * 7 * 6. Wait, because each card is drawn without replacement
Top 16 Negative Activations
Rollout 6, Act: -2.461
this case, the oracle TM can, given input x of length n, retrieve s_n from the oracle
Rollout 33, Act: -2.462
than k in the first k positions, which is equivalent to saying that the firstk positions cannot be a
Rollout 1, Act: -2.474
is d, I need to find a point y such that { sqrt(2)/d (x -
Rollout 93, Act: -2.480
4π/1007 = π/2 + π/(2*1007).
Rollout 15, Act: -2.489
over GF(2). Therefore, the original problem reduces to this.\n\nHence, The answer is that
Rollout 19, Act: -2.522
7 modk⇒83≡0 modk⇒k divides83. Since83 is prime
Rollout 70, Act: -2.524
(7)).\n\nCalculating that:\n\n(ln(2/7)) / (ln(10/
Rollout 22, Act: -2.540
62618≈1.2206. Total of that: 81.
Rollout 19, Act: -2.550
modk⇒ 83≡0modk⇒k divides83. Therefore, possible k:
Rollout 86, Act: -2.562
) - (1/4)sin(pi) + (1/32)sin(2pi
Rollout 90, Act: -2.567
43,960 +4,036=448, - 44
Rollout 32, Act: -2.659
+669\n\nTerm2 has +670 and -670ω\n\nTerm3
Rollout 34, Act: -2.679
} +9 e^{s}) = (4 -12) + (-2 e^{-s}
Rollout 50, Act: -2.685
)\n\n= \((\log_{10}2)^2 + \log_{10}1
Rollout 20, Act: -2.692
congruence mod 2021 is equivalent to the congruence mod 43 and
Rollout 74, Act: -2.853
= H_e + H_induced_N = H_e [1 + K_b cos2θ - K

Layer 46

GATE_PROJ

Top 16 Positive Activations
Rollout 52, Act: 3.182
\boxed{-2013}\n\n<|im_start|>answer\nTo determine the smallest yummy integer \( A \),
Rollout 58, Act: 3.074
2037171}\n\n<|im_start|>answer\nGiven an exam with \( k > 1 \
Rollout 93, Act: 3.064
{2^{2014}}}\n\n<|im_start|>answer\nTo compute the sum \(\sum_{k
Rollout 92, Act: 3.058
.78} diopters.\n\n<|im_start|>answer\nThe student is farsighted and uses glasses to
Rollout 32, Act: 3.038
\boxed{4018}\n\n<|im_start|>answer\nGiven the matrix \(\mathbf{M}
Rollout 84, Act: 3.030
Answer**\n\boxed{37}\n\n<|im_start|>answer\nLet the initial amount Gary purchased be 1 unit
Rollout 23, Act: 2.998
Final Answer**\n\boxed{B}\n\n<|im_start|>answer\nTo determine the activity of the alpha decay after
Rollout 70, Act: 2.992
**\n\boxed{315}\n\n<|im_start|>answer\nA positive integer \( N \) is called a
Rollout 35, Act: 2.971
**\n\boxed{418}\n\n<|im_start|>answer\nGiven a polynomial \( P(x) \) with
Rollout 46, Act: 2.969
\frac{dx}{dt}}\n$$\n\n<|im_start|>answer\nTo express \(\frac{d^{2
Rollout 25, Act: 2.969
}}$ to make contact with Earth.\n\n<|im_start|>answer\nTo determine how massive the asteroid 202
Rollout 9, Act: 2.965
Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides of length
Rollout 3, Act: 2.961
{169} m/s.\n\n<|im_start|>answer\nTo compute the mean molecular speed \( v \)
Rollout 80, Act: 2.942
Answer**\n\boxed{11}\n\n<|im_start|>answer\nTo determine the time scale \( t \) it
Rollout 41, Act: 2.936
\boxed{927}.\n\n<|im_start|>answer\nTo determine how many positive integers less than 1
Rollout 64, Act: 2.923
9} + 1\right)}\n\n<|im_start|>answer\nGiven a sequence of positive integers \(a_1
Top 16 Negative Activations
Rollout 22, Act: -4.040
speed is sqrt(8kT/(πm)). So the user is asking for the mean speed,
Rollout 3, Act: -4.057
^2 = 28,561. So sqrt(28,436)
Rollout 80, Act: -4.078
0^1 is 10^25. So if I have 92.64
Rollout 59, Act: -4.095
1/7 = 4/28.\n\nLet me verify that these sum up to 1
Rollout 61, Act: -4.109
3\n\n133:7*19\n\nNo common factors, so gcd(7*1
Rollout 2, Act: -4.111
)^{997} mod125\n\nBecause from k=3 to k=99
Rollout 0, Act: -4.139
,17,19 have exponent 1.\n\nSo primes 2: 18,
Rollout 80, Act: -4.145
10^25 g. Got it. Let me confirm:\n\n12.8 *
Rollout 2, Act: -4.157
k -1) �� -1 mod8.\n\nTherefore, for k=1: (2^
Rollout 17, Act: -4.192
48×10^-33 m3.\n\nTherefore, the entire |I|= (11
Rollout 31, Act: -4.198
quadratic: x=(1 ±√5)/2.\n\nThus, the roots of the cubic are x=-
Rollout 51, Act: -4.208
when �� is π. So the limits work. Therefore, I can split the integral into three parts
Rollout 2, Act: -4.212
P ��109 mod125.\n\nSo summary so far:\n\nP≡5 mod8
Rollout 89, Act: -4.220
3/5 * T = 100.\n\nTherefore, T = (100 *5
Rollout 94, Act: -4.251
2i)*(-2) = -4i\n\nYes, -4i equals -4i.
Rollout 22, Act: -4.405
_mean = sqrt(8kT/(πm))\n\nAlright, so first I need to find m,

UP_PROJ

Top 16 Positive Activations
Rollout 84, Act: 2.412
32 = 37 \).\n\nThe final answer is \(\boxed{37}\).<|im_end|>
Rollout 69, Act: 2.243
2 = 204\).\n\nThe final answer is:\n\[\n\boxed{204
Rollout 73, Act: 2.120
398\n\]\n\nThus, the final answer is:\n\[\n\boxed{398
Rollout 47, Act: 2.118
= 9\n \]\n\nThus, the value of \( N \) is \(\boxed{
Rollout 71, Act: 2.087
98\n\]\n\nThus, the final answer is:\n\[\n\boxed{98}\n
Rollout 83, Act: 2.069
144\n\]\n\nThus, the number of distinguishable acceptable arrangements is \(\boxed{
Rollout 88, Act: 2.044
100\).\n\nThus, the total number of weights defined for the entire activation output of the
Rollout 95, Act: 2.027
490}\) rational.\n\nThus, the number of segments \(\overline{P_kQ
Rollout 32, Act: 2.025
4 = 4018 \]\n\nThe sum of the entries of \(\mathbf{
Rollout 81, Act: 2.020
frac{510}{11}\).\n\nThe area of triangle \(ABC\) is \(\boxed
Rollout 84, Act: 2.020
4 - 4x\n\]\n\n\[\n9 - 4 = 36x -
Rollout 36, Act: 2.018
192\n\]\n\nThus, the final answer is \(\boxed{192}\).
Rollout 32, Act: 2.002
4 = 4018 \]\n\nThe sum of the entries of \(\mathbf{M
Rollout 4, Act: 1.973
0 + 3000} = \frac{6000000}{5
Rollout 73, Act: 1.966
00 = 398\n\]\n\nThus, the final answer is:\n\[\n\boxed
Rollout 60, Act: 1.961
, \text{Hz}\n\]\n\nThe beat frequency is the difference between the reflected frequency \( f''
Top 16 Negative Activations
Rollout 79, Act: -3.002
from these letters include:\n\n- "Hernias" plus some others. No. Maybe "Chairmen
Rollout 49, Act: -3.003
sets) is called a "perfectly normal" space? Not exactly. Perfectly normal means that every
Rollout 11, Act: -3.003
number is real, its complex conjugate is itself. So for p to be real, p = conjug
Rollout 0, Act: -3.018
, 3^8, 5^4, etc., up to the primes less than or equal
Rollout 12, Act: -3.030
2?\n\nAlternatively, substitution φ where t = cosθ? Let me think. For if t is from
Rollout 81, Act: -3.034
atic formula: h = [ -70 ± sqrt(70^2 - 4*1
Rollout 62, Act: -3.043
's famous polynomial n^2 +n +41, which generates primes for n=0 to
Rollout 54, Act: -3.053
start with ethanethiolate (most reactive) > hydroxide > propionate? No,
Rollout 93, Act: -3.071
14}/sqrt(pi*1007)} } by the approximation of central binomial coefficient:
Rollout 12, Act: -3.074
diverging.\n\nAlternatively, with substitution t = sinθ, but seems not applicable here. Wait, let
Rollout 91, Act: -3.094
2 - AC2)/(4 * area of triangle). Wait, not exactly. Let's recall that in
Rollout 62, Act: -3.123
-m +11. From m=1 to 10, primes. Let me check m
Rollout 68, Act: -3.153
's: k=8 (valid) and k in k>8 and k<0 (but need
Rollout 35, Act: -3.154
solutions to this quadratic are n=(41 ± sqrt(1681 -1672
Rollout 98, Act: -3.202
0.9 solar masses would be a G-type or maybe K-type. So a 0.
Rollout 34, Act: -3.279
both equations are Volterra integral equations of the second kind, right? The kernel is \( e^{-2

DOWN_PROJ

Top 16 Positive Activations
Rollout 96, Act: 4.978
06*169= Let's compute:\n\n4106*100=4
Rollout 4, Act: 4.800
Let me calculate 4.327 divided by 1200.\n\nDivide by
Rollout 4, Act: 4.714
compute that more accurately.\n\n4.327 / 1200. Let me calculate
Rollout 96, Act: 4.388
:\n\n-4106*169= Let's compute:\n\n4106*1
Rollout 81, Act: 4.378
; to compute 44*561:\n\nFirst, think 561 * 4
Rollout 22, Act: 4.311
8.7768 �� 0.006333 �� 1
Rollout 36, Act: 4.264
,624*30,624:\n\n30,000*30
Rollout 22, Act: 4.193
2 * 273 = let's compute:\n\n66.512 * 20
Rollout 3, Act: 4.175
=66.48*298= same approach: 66.48*
Rollout 92, Act: 4.137
37*123. Let me compute:\n\n37*120 = 44
Rollout 36, Act: 4.135
584):\n\n147456 - 55584 = Let's compute
Rollout 3, Act: 4.090
6.512 * 298:\n\nLet me compute this multiplication.\n\n66.5
Rollout 80, Act: 4.084
.8 * 7.238 is 92.6464, yeah,
Rollout 4, Act: 3.968
0 / 5025. \n\nDividing 6,060,150
Rollout 22, Act: 3.957
6.512 * 273 = let's compute:\n\n66.512
Rollout 17, Act: 3.951
42/1395.18≈0.02538. Yes.\n\n
Top 16 Negative Activations
Rollout 56, Act: -3.063
area S_n=K_n= (sum from k=1 to n+1 of 1/
Rollout 89, Act: -3.073
333... Therefore, 2 parts girls: 66.666...\n\nAgain same
Rollout 82, Act: -3.095
codeword is length2, sum is2*(all probs)=2*69/69
Rollout 24, Act: -3.130
C^{\prime} \subseteq C$, we have $R^{\prime}=R$ and $
Rollout 56, Act: -3.142
).\n\nDistance formula:\n\nA3A1 = sqrt[(1 - 0)^2 + (0 -
Rollout 73, Act: -3.162
with n=1: sum should be 1*2 =2.\n\nPlug in n=1:
Rollout 18, Act: -3.168
+5k))/5 = (-3 -12 -20k)/5 = (-15
Rollout 67, Act: -3.169
Then given that:\n\nFor n=1: A(1)^2 + B(1) + C
Rollout 24, Act: -3.212
subseteq C$, we have $R^{\prime}=R$ and $C^{\prime}=C
Rollout 45, Act: -3.248
Similarly, f(2)=20:\n\nk(8) +m(4) +n(
Rollout 93, Act: -3.249
the sum is (C(2014,0)*1007)/2^{2
Rollout 45, Act: -3.293
f(1)=1, which gives:\n\nk(1)^3 +m(1)^2 +
Rollout 44, Act: -3.354
2 - V1) = 2*(2 -1)=2 L·atm. So indeed
Rollout 24, Act: -3.470
\subseteq C$, we have $R^{\prime}=R$ and $C^{\prime}=
Rollout 14, Act: -3.542
, r=4, since 1/2 +1/4 +1/4=1.\n\n
Rollout 95, Act: -3.614
1}) = (k - (k-1))*(Area ABC)/2450 = Area ABC

Layer 47

GATE_PROJ

Top 16 Positive Activations
Rollout 68, Act: 3.173
)$ has exactly one real solution.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the
Rollout 97, Act: 3.140
of $\mathrm{Y}$?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I have
Rollout 47, Act: 3.138
the value of $N$?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so there's this problem about
Rollout 91, Act: 3.130
- \cot C|.$\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this geometry problem
Rollout 66, Act: 3.117
$ . Find $p$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through
Rollout 86, Act: 3.112
}}f(x)\ dx.$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, so the problem is asking for
Rollout 24, Act: 3.107
contain the same number of rows.\n<|im_start|>assistant\n<|im_start|>think\nAlright, let me start by trying to
Rollout 36, Act: 3.092
}$ . Find $AP.$\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure out
Rollout 40, Act: 3.061
p| + |q|$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's tackle this problem step
Rollout 30, Act: 3.054
+3}}\right)$.\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I've got this problem
Rollout 17, Act: 3.045
}|$ (in SI units).\n<|im_start|>assistant\n<|im_start|>think\nAlright, let me try to tackle this
Rollout 61, Act: 3.042
Find $p+q.$\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's see. So the
Rollout 1, Act: 3.042
\mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show that
Rollout 31, Act: 3.032
possible values of $y$.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's try to tackle this
Rollout 32, Act: 3.027
2009}.$\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to find the
Rollout 72, Act: 3.026
$ \n1. $ \n<|im_start|>assistant\n<|im_start|>think\nAlright, let's tackle this problem:
Top 16 Negative Activations
Rollout 89, Act: -4.050
00 multiplied by 5 is 500, divided by 3 is approximately 16
Rollout 20, Act: -4.055
6 divides 966? 46 divides into 966? 46 *
Rollout 8, Act: -4.058
��240= 126. So 31/126.\n\nWait
Rollout 3, Act: -4.060
). Let's see, 168^2 = 28,224.
Rollout 79, Act: -4.064
8-letter word followed by a 4-letter word. Let's try to find possible combinations. Maybe
Rollout 59, Act: -4.071
divided by 784. Reduction: GCD(246,784).\n\nPrime
Rollout 76, Act: -4.073
17 and 512 are coprime.\n\nSince 512 is 2^
Rollout 79, Act: -4.085
to get the answer. Therefore, solving an anagram of "minorchangesin" (considering "
Rollout 19, Act: -4.115
cubes up to2:1+8=9. 9 divided by7 gives remainder9-7
Rollout 3, Act: -4.131
,820.576.\n\nDivide that by π * 0.222
Rollout 8, Act: -4.141
126. Then check if they can be simplified. Since 31 is prime. 1
Rollout 19, Act: -4.152
cubes1+8+27=36. Divided byn+5=8, remainder
Rollout 81, Act: -4.241
= let's check: 172^2=29584 because 17
Rollout 16, Act: -4.276
0 and 1953 are coprime. As previous calculation using gcd steps shows that gcd
Rollout 76, Act: -4.300
2. But 417 +512 would be 929.\n\nHence,
Rollout 59, Act: -4.340
2. Then 123 +392= 515.\n\nBut let's check

UP_PROJ

Top 16 Positive Activations
Rollout 66, Act: 4.459
is odd, then m would be odd + even? Wait, m=12n2 + n
Rollout 70, Act: 4.293
331 +16=331? Wait, wait N=35*9 +
Rollout 32, Act: 4.264
+0 + (−ω2 from S) ?? Wait no. Let me recomputethe previous step
Rollout 93, Act: 4.256
=1: 3/2, not. Wait, for m=1, sum is (1
Rollout 96, Act: 4.208
7+6+9+3=34?\n\nWait, wait:\n\n5+8=13
Rollout 71, Act: 4.180
8. Is 2008 a square? sqrt(2008)=~44
Rollout 33, Act: 4.171
.\n\nThus, total:3+2=3? Wait, the entire permutations:\n\nTotal n=3
Rollout 91, Act: 4.169
2t2 -2t +1}/2 ?\n\nWait, again similar steps. Let's go through
Rollout 96, Act: 4.161
+1134)/169.\n\nWait, in previous calculations, OB_x was (-8
Rollout 95, Act: 4.151
750=2*3675?\n\n 3675*2=73
Rollout 22, Act: 4.136
732.4 at numerator, is that? No, numerator is 18,15
Rollout 48, Act: 4.119
sum, it's F B 1 2? Wait, let's compute each digit:\n\nRightmost
Rollout 39, Act: 4.091
10^3)/9 - (3). Wait, if we follow the formula as before, sum
Rollout 20, Act: 4.083
7, 46 divides 966? 46 divides into 966?
Rollout 0, Act: 4.053
4,7:2, which are all even? Wait, 18,8,4,
Rollout 66, Act: 4.050
odd, then m would be odd + even? Wait, m=12n2 + n.
Top 16 Negative Activations
Rollout 26, Act: -2.330
answer\nTo find the standard molar enthalpy of formation of $\mathrm{HgO}(
Rollout 57, Act: -2.341
think\nOkay, so I need to show that if we define a neighborhood basis around a point x on
Rollout 15, Act: -2.353
the behavior of the system using linear algebra over the field \( \text{GF}(2) \).
Rollout 77, Act: -2.358
}{4} \cdot f(10, 4) \)\n- \( f(10
Rollout 87, Act: -2.360
boxed{ \dfrac{3 \rho a^2}{4 \pi \epsilon_0}
Rollout 46, Act: -2.362
dt}}\n\]\nand\n\[\n\boxed{\frac{d^{2} y}{d t
Rollout 10, Act: -2.384
- A position is losing if all possible moves lead to winning positions.\n - Through analysis, we
Rollout 46, Act: -2.396
}{\partial y} - 2 \omega \frac{dx}{dt}}\n\]<|im_end|>
Rollout 74, Act: -2.414
\theta = K_b \sin \theta + K_s \cos \theta} \]<|im_end|>
Rollout 70, Act: -2.415
when read as a base-10 number, equal twice \( N \). To find the largest such
Rollout 82, Act: -2.422
0\n- \( \frac{12}{69} \): 90\n- \(
Rollout 67, Act: -2.462
2 + 492(8) + 1020\n\]\n\[\nc
Rollout 93, Act: -2.486
1 to 1007 of cosine raised to the 2014th power of (
Rollout 95, Act: -2.524
the triangle into 2450 regions of equal area. We need to find the number of segments
Rollout 0, Act: -2.544
}\) and calculate the product of the resulting numerator and denominator. We need to determine how many such rational
Rollout 11, Act: -2.564
{b}{c} + \frac{c}{a}\), given that both \(p\) and

DOWN_PROJ

Top 16 Positive Activations
Rollout 41, Act: 3.574
a, three-digit numbers. So 3 per a, so 9*3=27.\n\n
Rollout 56, Act: 3.321
this implies that each next term is multiplied by (something) following the formula.\n\nBut this pattern is not
Rollout 80, Act: 3.271
Specific heat: 0.4 J/gK. Delta T=1. So Q=9.
Rollout 4, Act: 3.224
0.52 +0.52 +something)%? Wait, no—wait:\n\nWait,
Rollout 80, Act: 3.212
.265e25 *0.4 =3.706e25 J
Rollout 81, Act: 3.189
. Wait, leading coefficients differ. Huh. What's causing this?\n\nWait, in the coordinate system
Rollout 41, Act: 3.168
, three-digit numbers. So 3 per a, so 9*3=27.\n\nTherefore
Rollout 71, Act: 3.165
Now, looking for c such that 3c2 + k2 = 8044.
Rollout 45, Act: 3.153
: c = 0\n\nSo c = 0. Then from Equation 1: 2a
Rollout 70, Act: 3.129
2*2400=4800≠6666, so that doesn't
Rollout 70, Act: 3.125
but 2*342=684≠666. So no.\n\nOr6
Rollout 13, Act: 3.101
^2.\n\nPerhaps set a = 1.25, b=0.0044
Rollout 65, Act: 3.099
. Therefore x / a = 1, so x = a, which doesn't make sense. Alternatively
Rollout 51, Act: 3.091
Therefore, perhaps we can shift the variable y = x - ��, so dy = dx, and
Rollout 70, Act: 3.049
9, which is less than 343? Wait 499 is higher than 3
Rollout 85, Act: 3.021
the direction steps:\n\nIf f(x,y+1) - f(x,y) = 1 and f
Top 16 Negative Activations
Rollout 83, Act: -3.198
Also, arrangements that can be rotated into each other are considered the same, so we have to account for
Rollout 13, Act: -3.213
1.12\nx1.12\n--------\n224 (1.12 *
Rollout 99, Act: -3.224
3.199 +3≈0.505 -3.199 +3
Rollout 22, Act: -3.226
that they didn't say the temperature, this problem is incomplete.\n\nWait, but perhaps in the process of
Rollout 17, Act: -3.256
·m2/s2) ) / T? Wait, maybe it's simpler to note that units check
Rollout 31, Act: -3.270
1+\sqrt{5}}{2}}\), and \(\boxed{\frac{1-\sqrt
Rollout 73, Act: -3.282
=1, 4-1=3, 4-2=2. Sum is 1
Rollout 99, Act: -3.317
) = f(0.5) �� 2.19897\n\nf(x
Rollout 42, Act: -3.326
cohols are treated with acid. Phosphoric acid is a strong acid, and in such conditions,
Rollout 0, Act: -3.344
2,3,5:\n\nThe 8 assignments:\n\n1. Assign none to a => a=1
Rollout 62, Act: -3.511
that has at least four prime factors, right? They don't have to be distinct, so even if
Rollout 76, Act: -3.620
), (1,2), (2,1), or (2,2) if we label the
Rollout 99, Act: -3.811
1573) �� 0.334 (since 10^0.
Rollout 22, Act: -3.827
of hydrogen gas (H2) at room temperature, I guess? The question doesn't specify the temperature
Rollout 76, Act: -4.190
7 factors 3*139 [since 1+4+1+7=13
Rollout 99, Act: -4.331
1575) �� 0.334 (since 10^0.

Layer 48

GATE_PROJ

Top 16 Positive Activations
Rollout 14, Act: 4.709
) \, dx \right) }\n\n<|im_start|>answer\nTo prove the inequality \n\n\[\n\left(\
Rollout 90, Act: 4.667
0891928}\n\n<|im_start|>answer\nTo determine the smallest positive integer \( A \)
Rollout 9, Act: 4.624
Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides of length
Rollout 81, Act: 4.591
{510}{11}}\n\n<|im_start|>answer\nGiven triangle \(ABC\) with \(\tan \
Rollout 95, Act: 4.572
Final Answer**\n\boxed{0}\n\n<|im_start|>answer\nTriangle \(ABC\) has side lengths \(AB =
Rollout 91, Act: 4.556
Final Answer**\n\boxed{2}\n\n<|im_start|>answer\nGiven triangle \(ABC\) with median \(AD\)
Rollout 0, Act: 4.533
**\n\boxed{128}\n\n<|im_start|>answer\nGiven a rational number between 0 and 1
Rollout 1, Act: 4.524
y \text{ exists as described}}\n\n<|im_start|>answer\nGiven an infinite-dimensional Hilbert space \( \math
Rollout 35, Act: 4.516
**\n\boxed{418}\n\n<|im_start|>answer\nGiven a polynomial \( P(x) \) with
Rollout 16, Act: 4.503
{789953}\n\n<|im_start|>answer\nTo determine the expected number of days it will take
Rollout 2, Act: 4.468
**\n\boxed{109}\n\n<|im_start|>answer\nTo find the remainder when \(9 \times
Rollout 31, Act: 4.465
{5}}{2}}\).\n\n<|im_start|>answer\nGiven real numbers \( x \) and \( y
Rollout 49, Act: 4.463
X \text{ is countable}}\n\n<|im_start|>answer\nTo prove that if every subspace of a Haus
Rollout 60, Act: 4.463
Final Answer**\n\boxed{6}\n\n<|im_start|>answer\nA stationary source emits sound of frequency \( f_
Rollout 25, Act: 4.462
}}$ to make contact with Earth.\n\n<|im_start|>answer\nTo determine how massive the asteroid 202
Rollout 36, Act: 4.462
**\n\boxed{192}\n\n<|im_start|>answer\nGiven a cube with edges \(\overline{
Top 16 Negative Activations
Rollout 79, Act: -4.779
changes in investigators' facility" is the wordplay. But the way it's phrased is "
Rollout 87, Act: -4.782
set u = y, v = z. Not helpful. Alternatively, non-dimensionalize the variables by defining
Rollout 79, Act: -4.790
" as "knighted". Doesn't make sense.\n\nAnother approach: Think of abbreviations or hy
Rollout 48, Act: -4.795
xB612 (wrong way in previous step). Wait no, the third word is 0xB
Rollout 11, Act: -4.812
y/z, since z is conjugate of c. Hmmm, perhaps requiring relationships like a/b =
Rollout 82, Act: -4.814
Then pair the next two, etc. Let's see:\n\nGroup1: p1, p5\n\n
Rollout 59, Act: -4.819
+4/28 z3\n\nCombine like terms:\n\nConstant term: 3/28\n\n
Rollout 93, Act: -4.888
=2*(1/2) + 0? Not sure. When m=2:1 (
Rollout 13, Act: -4.905
5. Alternatively, perhaps use the standard multiplication grid. But this might take longer.\n\nAlternatively, another method
Rollout 91, Act: -4.929
computed before may better, even though more steps. Let's proceed.\n\nCalculating angle at C:\n\nVectors
Rollout 4, Act: -4.939
Wait, this is another way of writing the terms. Maybe this will increase precision?\n\nWait, or rather
Rollout 73, Act: -4.970
, that seems circular. Maybe this method isn't working.\n\nWait, perhaps another approach. Wait, use
Rollout 4, Act: -4.975
49e-3*sqrt(10). Because sqrt(10)=3.16
Rollout 36, Act: -4.979
,000 + (-864)\n\nWait,\n\nWait, better to use equation(d): s
Rollout 41, Act: -4.991
then sum them up for all lengths. Let's see.\n\nDefine S as the total across all lengths (
Rollout 11, Act: -5.111
variables is the product of conjugates of the others. For instance, maybe c = \overline{

UP_PROJ

Top 16 Positive Activations
Rollout 63, Act: 3.595
, \( pq = 0 \), so the base case is satisfied.\n\nTherefore, it seems that \(
Rollout 33, Act: 3.593
(n) be the number of permutations ofn elements satisfying that for1 <=k <=n-1,
Rollout 73, Act: 3.588
00=398.\n\nTherefore, the remainder should be 398.\n\nWait, but
Rollout 0, Act: 3.574
a share no primes with b. Therefore, the coprime pairs (a,b) with a*b
Rollout 58, Act: 3.531
1 * P(0). Therefore, each term in the sum for part i is i * something...
Rollout 30, Act: 3.524
spaced by two indices. Since it's an infinite sum, maybe it telescopes? If that's the
Rollout 41, Act: 3.519
are allowed. Therefore, total two-digit numbers that meet the conditions would be all 90 numbers,
Rollout 45, Act: 3.518
but f(x) is cubic. For part (b), perhaps it's only for non-negative integers x
Rollout 72, Act: 3.504
notation. If unsure, but knowing the problem requires writing all even positive integers, the safest is to state
Rollout 15, Act: 3.498
4. Therefore, according to the linear algebra model, any initial state should die out after at most
Rollout 68, Act: 3.492
no real solutions.\n\nBut we need to remember that these solutions must also satisfy x > -2 and k
Rollout 76, Act: 3.481
95. Therefore, total colorings with at least one 2x2 red square is 9
Rollout 89, Act: 3.474
66.666...\n\nAgain same answer.\n\nBut again, that gives us a fractional answer
Rollout 89, Act: 3.470
's precise. So maybe this problem is a trick question, but typically though, problems would have the fractional
Rollout 13, Act: 3.468
is 187.12 which seems reasonable.\n\nSo my conclusion is that the extra amount Jo
Rollout 16, Act: 3.467
answer is supposed to be m and n coprime, so m=7880 and n
Top 16 Negative Activations
Rollout 78, Act: -2.632
{v_e - 20}{60} \):\n \[\n L = \frac
Rollout 95, Act: -2.632
5= sqrt(671*3)/5= sqrt(2013)/5.\n\n
Rollout 64, Act: -2.693
(2^{2009} -1) + a1(2 - 2^{2
Rollout 7, Act: -2.712
P =79x - (A +5)x\n\n-3P=(74 -A)
Rollout 13, Act: -2.713
, b = 0.1255\n\nBut that's not helpful. Let me compute
Rollout 53, Act: -2.721
( 3d - 2d ) /6 = d/6\n\nTherefore, PD distance is
Rollout 40, Act: -2.737
{2} + 1))(1 + i) \]\nExpanding the product:\n\[ (1
Rollout 95, Act: -2.754
3 s - 9t) - 9t??? Wait.\n\nWait, hold on. Let me
Rollout 86, Act: -2.756
/2)( - (-1 -1 ) /2 )= - (1/2)( - (-
Rollout 87, Act: -2.764
(1 + r2))) * 2r ] = r / sqrt(1 + r2)\n\n
Rollout 78, Act: -2.802
frac{v_e - 20}{60} \):\n \[\n L = \
Rollout 81, Act: -2.853
70 \pm 172}{22} \implies h = \frac{1
Rollout 11, Act: -2.862
a bc =1, so ab*c =a. So a b c =1. Hence, multiplying
Rollout 71, Act: -2.944
(-a - b)a + (-a - b)b = ab - a^2 - ab - ab
Rollout 30, Act: -2.949
(n+3)-1} +1} ) = ?\n\nWait, no. Let's recall the
Rollout 7, Act: -3.121
= (56 -54)N/3 ?\n\nWait, wait:\n\nWait 18N

DOWN_PROJ

Top 16 Positive Activations
Rollout 96, Act: 5.070
2 to 132*(169/169)= 22308
Rollout 61, Act: 4.463
Find $p+q.$\n<|im_start|>assistant\n<|im_start|>think\nAlright, let's see. So the
Rollout 89, Act: 4.422
many girls are in the class?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me see here. So
Rollout 85, Act: 4.422
,2024)$.\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm trying to solve
Rollout 16, Act: 4.419
/2)^6 = 1 - (1/2)^6. Wait, that's actually correct
Rollout 84, Act: 4.406
. Find $m+n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this
Rollout 20, Act: 4.400
factorization of $n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the
Rollout 73, Act: 4.389
$1000$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to solve this
Rollout 35, Act: 4.379
\cdot n_2.$\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find the
Rollout 59, Act: 4.379
Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so the problem is about Jackie
Rollout 76, Act: 4.361
Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to find the probability
Rollout 2, Act: 4.358
$1000$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this problem here
Rollout 8, Act: 4.332
Find $m + n$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. We need
Rollout 30, Act: 4.325
+3}}\right)$.\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I've got this problem
Rollout 55, Act: 4.323
2018}.$\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I have this matrix A
Rollout 56, Act: 4.315
an area of $K$?\n<|im_start|>assistant\n<|im_start|>think\nAlright, so Wesyu is building this
Top 16 Negative Activations
Rollout 36, Act: -3.003
456 - 55584 = Let's compute 147456
Rollout 21, Act: -3.006
integer. For example, floor(x) is surjective onto Z. Similarly, certain step functions. But
Rollout 14, Act: -3.020
dx - d \int x^2 dx = c*( ( (c/d)^2 ) /2
Rollout 37, Act: -3.032
but the problem states that the a_i are integers, but not necessarily distinct, and the k_i are
Rollout 89, Act: -3.037
boys? But in real life, you can't have a fraction of a person. Therefore, the total
Rollout 14, Act: -3.041
$-\mu K$ is $-\mu x$. So setting the functional derivative to zero gives:\n\n$$
Rollout 2, Act: -3.041
9: 891*999=891*(1000 -1
Rollout 52, Act: -3.063
is correct. Wait, the problem mentions that A has to be part of the sequence which adds to
Rollout 21, Act: -3.092
) = k for integer k, but variables can take any real number. However, functional equations that are
Rollout 36, Act: -3.100
,362.666..., which is not a whole number. Maybe there was a calculation
Rollout 2, Act: -3.210
9= 891*999= let's compute: 891*(1
Rollout 15, Act: -3.233
whether that single neighbor is on. For all other lamps, since they have two neighbors, their next state
Rollout 37, Act: -3.238
the problem states that the a_i are integers, but not necessarily distinct, and the k_i are positive
Rollout 67, Act: -3.503
2: q= -72 / d.\n\nTherefore, substituting into equation 3:\n\na
Rollout 23, Act: -3.783
.073... min−1 – that's correct. \n\nTherefore, λ_total ��
Rollout 71, Act: -3.896
11). sqrt(2011) is approx 44.86...\n\nSo primes

Layer 49

GATE_PROJ

Top 16 Positive Activations
Rollout 82, Act: 5.598
len2).\n\nAnd 15 is0, which is len1.\n\nSo Wait, actual codew
Rollout 6, Act: 5.365
then O(n^k / log n), but this is still not polynomial per length. Wait.\n\nAlternatively
Rollout 75, Act: 5.351
Therefore, S=8N -5. But this may complicate things. Wait, no, perhaps
Rollout 70, Act: 5.349
so 2*49=98 (which is two-digit) to 2*34
Rollout 38, Act: 5.166
, which would mean x =2L, but this would make point D=(-2L, W),
Rollout 6, Act: 5.148
have p(n) strings for each n, but this might not be sparse if p(n) is a
Rollout 75, Act: 5.146
? Only 1 + 8, which 8 is not a square; 4 + 5
Rollout 15, Act: 5.059
which is 2^2 -1: no, since n=3 was bad. Maybe another pattern
Rollout 10, Act: 5.045
: subtract16=10 (L) which is previous, so from 26 can go
Rollout 28, Act: 5.039
y +a2b2=0\n\nBut this gives:\n\n(2b x +2a y
Rollout 75, Act: 5.037
6+16=32 already. But that's 16+16+0,
Rollout 49, Act: 5.012
-compact. Therefore, hereditary σ-compactness is a stronger condition.\n\nSo in our case,
Rollout 11, Act: 5.008
0. Then p= -2. But maybe this is invalid.\n\nGiven previous problem maybe David Altizio
Rollout 28, Act: 4.980
\n\nWhich is exactly what we started with. So this transformation doesn't help. Alternatively, equation of the
Rollout 38, Act: 4.980
=2*(x/2)=x. So no, it's L=x/2. But then
Rollout 15, Act: 4.975
. All off. So within one step! But this seems contradictory. Wait, maybe a mistake.\n\nWait
Top 16 Negative Activations
Rollout 46, Act: -7.009
}+y^{2}\right)$.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have to
Rollout 25, Act: -7.010
$r$ is the impact parameter.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure
Rollout 87, Act: -7.014
0}$, and any necessary numerical constants.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to
Rollout 29, Act: -7.020
P < \rho gH$ \n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure
Rollout 57, Act: -7.021
. \n \n*A. Csaszar*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show
Rollout 48, Act: -7.033
0 . Does the header contains error?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I have this IPv
Rollout 54, Act: -7.035
3, B = 2\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this question
Rollout 3, Act: -7.046
(Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to calculate the
Rollout 22, Act: -7.060
(H2) in m/s\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need to calculate
Rollout 98, Act: -7.082
4\n(D) 2\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to figure
Rollout 92, Act: -7.091
refractive power of his glasses lenses?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure
Rollout 58, Act: -7.114
Proposed by Yannick Yao*\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. So
Rollout 23, Act: -7.154
7.555 Bq\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this problem
Rollout 94, Act: -7.175
c,d\n(D) a,b\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to determine
Rollout 43, Act: -7.264
left|0\right\rangle\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure
Rollout 42, Act: -7.302
4]oct-5-ene\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to figure

UP_PROJ

Top 16 Positive Activations
Rollout 95, Act: 4.122
and Qk.\n\nSince Pk is on AB, which goes from A(0,0) to
Rollout 95, Act: 4.059
1/3). So we can parametrize Pk and Qk with parameter t, where t is
Rollout 72, Act: 3.753
no. The question requires "Find all numbers n...", hence the answer is all even integers n ≥2
Rollout 32, Act: 3.685
3, and the leftover terms.\n\nNumber of cycles: 2009 divided by 3 gives
Rollout 95, Act: 3.628
0) to (9,0), so APk is (9k/2450,
Rollout 95, Act: 3.616
t is the scaling factor, then coordinates of Pk would be (9t, 0), and
Rollout 95, Act: 3.607
divides AB into 2450 equal parts, APk = (9/2450
Rollout 95, Act: 3.594
recast this.\n\nPoint Pk is along AB. AB from (0,0) to (9
Rollout 6, Act: 3.592
S is sparse.\n\nThen, the oracle machine M, when given <x,i>, encodes i and
Rollout 95, Act: 3.586
is the scaling factor, then coordinates of Pk would be (9t, 0), and coordinates
Rollout 6, Act: 3.584
bit of α_n. Then, the machine M, given x of length n, needs to get the
Rollout 61, Act: 3.581
for d=9: k = m +9, m can go from 0 to 39
Rollout 81, Act: 3.565
's compute the slopes. \n\nThe slope of AB: since A is (0,h) and B is
Rollout 14, Act: 3.550
.\n\nFor I1=∫0^A f dx:\n\n applyCauchy-Schwarz: I
Rollout 95, Act: 3.547
things, but perhaps let's parametrize points Pk and Qk.\n\nSince Pk is on AB
Rollout 6, Act: 3.541
sparse.\n\nTo retrieve α_n, the oracle machine, given input x of length n, would need to
Top 16 Negative Activations
Rollout 70, Act: -1.899
2*343=1000 -686=314.\n\ni=
Rollout 36, Act: -1.912
728 + 37728 + 16128)s2 = (
Rollout 70, Act: -1.926
, 500 <=N <=4999.5. But N is also a four
Rollout 23, Act: -1.926
.88≈0.6931/5.88≈0.117
Rollout 96, Act: -1.932
169\n\n= (4563 +1134)/169\n\n=
Rollout 5, Act: -1.940
3b + 75 = 2b + 150 \implies b =
Rollout 96, Act: -1.944
)/169= (4563 +1134)/169=5
Rollout 93, Act: -1.949
2 /1007 �� 0.00987, so e^{-0
Rollout 4, Act: -1.953
} = \sqrt{12.96 + 5.76} = \sqrt{
Rollout 67, Act: -1.962
2 +32*(-6)=540 -192=348. Correct.\n\n
Rollout 16, Act: -1.967
864; 64*31=1984. 9864
Rollout 20, Act: -1.984
7,46)=47*46=2162.\n\nTherefore, n must be
Rollout 38, Act: -2.000
riminant=390,625 −112,896=27
Rollout 39, Act: -2.004
9)/9 = (10000 -2899)/9 = (71
Rollout 96, Act: -2.018
9.\n\nSo x=(22308 -13608)/169 /
Rollout 93, Act: -2.034
1007 �� 0.00987, so e^{-0.0

DOWN_PROJ

Top 16 Positive Activations
Rollout 99, Act: 4.860
)2 �� 0.6389\n\n4 a c = 4*(-0.
Rollout 90, Act: 4.732
=14,126. Total=403,600 +141
Rollout 4, Act: 4.674
3*25.2004=603*25 +603*
Rollout 68, Act: 4.472
me think how to approach this. Alright, logarithms... the equation involves logs, so first I should
Rollout 76, Act: 4.462
2x2 entirely red squares. It might be easier to compute the number of colorings that have
Rollout 36, Act: 4.346
0 +3,136=3,240,000+201
Rollout 62, Act: 4.194
11*97.\n\nThus, these happen when m is 11k or 1
Rollout 99, Act: 4.193
5)2 �� 0.6389\n\n4 a c = 4*(-0
Rollout 4, Act: 4.181
004e-12 �� 603*25.2004
Rollout 80, Act: 4.166
inos pass through it? Since the neutrinos are coming from the Sun, the flux at Earth's surface
Rollout 35, Act: 4.154
is an integer. But since a and b are separated by 7, we can possibly enumerate all pairs
Rollout 90, Act: 4.154
1928 is a four-digit number? 1,928. Yes. So
Rollout 13, Act: 4.146
Adding them together: 1.0609 + 0.063654
Rollout 48, Act: 4.141
numbers, when adding, if there's a carry beyond 16 bits, we need to add that
Rollout 87, Act: 4.139
. Let me see. Integration in three dimensions can sometimes be simplified by using spherical coordinates, but here the
Rollout 43, Act: 4.122
is the first qubit being |0⟩. That means this is a bit different from the standard C
Top 16 Negative Activations
Rollout 95, Act: -3.782
)) =\n\n( (25/3 -9)k/2450, sqrt(6
Rollout 56, Act: -3.841
x = 0 + t*(1 - 0) = t\n\ny = (-√3/
Rollout 7, Act: -3.843
this math problem about participants in a test. Let me first try to understand what's being asked here.\n\n
Rollout 86, Act: -3.867
1/2*(-1/2*(-1 -1 ))= -1/2*(-1/2
Rollout 7, Act: -3.871
/3) = (75*2)/3 *N=150/3 *N
Rollout 93, Act: -3.928
2014} C(2014,j) e^{iθ(201
Rollout 56, Act: -3.934
= (1/(2^{2} - 2)) A2A0 which is 1/(
Rollout 28, Act: -3.961
6(X2 - (2√2 /3) X ) +2Y2 +1=0
Rollout 45, Act: -3.975
) + 1 + a x^2 + a x + b x + c x + c +
Rollout 28, Act: -4.027
6(X2 - (4√2 /6) X ) +2Y2 +1=
Rollout 40, Act: -4.029
coefficients (-5 +5(√2 +2 ))i = (5√2 +5)i
Rollout 21, Act: -4.151
(x) = cx, so c(cx)=cx implies c2x = cx for all x. So
Rollout 36, Act: -4.155
00/s + 7200/s) x = (25200/s)x
Rollout 28, Act: -4.364
6(X2 - (4√2 /6) X ) +2Y2 +1=0
Rollout 7, Act: -4.412
5)x\n\n-3P=(74 -A) x.\n\nWait, substitute P=2N
Rollout 30, Act: -4.936
) - (b_{n+2} +1) = a_{n+1} b_{

Layer 50

GATE_PROJ

Top 16 Positive Activations
Rollout 82, Act: 2.673
0 +16+16)/26≈42/26≈1.61
Rollout 41, Act: 2.208
\)\n - Total for 4-digit numbers: \(9 + 504 + 6
Rollout 41, Act: 2.206
\)\n - Total for 3-digit numbers: \(9 + 216 + 2
Rollout 84, Act: 2.206
36x - 4x\n\]\n\n\[\n5 = 32x\n\
Rollout 72, Act: 2.204
\), and \( -1 \). Their sum is \( 1 + 1 + (-1)
Rollout 72, Act: 2.187
\( 1 + 1 + (-1) = 1 \).\n - This construction works for
Rollout 52, Act: 2.160
2} = 2014\n\]\nRearranging this, we get:\n\
Rollout 50, Act: 2.138
}{10} \right)^k\n\]\n\nSolving this gives:\n\n\[\nk = \
Rollout 97, Act: 2.112
be \(y\). Then, the total pressure is given by:\n \[\n P_{\
Rollout 4, Act: 2.087
. The equivalent capacitance for capacitors in series is given by:\n\n\[\nC_{\text{
Rollout 93, Act: 2.071
014}{2014} = 1\), we get:\n \[\n
Rollout 84, Act: 2.066
x = 4 - 4x\n\]\n\n\[\n9 - 4 = 36
Rollout 2, Act: 2.043
times (-1)^{997} \equiv 1 \times 3 \times (-1
Rollout 76, Act: 2.035
^2 = 4\) colorings. Total: \(2 \times 4 = 8\
Rollout 76, Act: 2.031
8\).\n - Combined total for overlapping pairs: \(32 + 8 = 40
Rollout 82, Act: 2.010
0 +16 +16)/26 = 42/26 ��1.
Top 16 Negative Activations
Rollout 11, Act: -5.575
some specific curves? Not sure.\n\nWait, another thought. Since abc = 1, take logarithms
Rollout 30, Act: -5.578
,211 /4,058,210.\n\nTherefore, the reduced form would
Rollout 88, Act: -5.579
0. But let's check.\n\nWait, another point: sometimes when people ask for the number of "
Rollout 3, Act: -5.594
m/s. But need to specify.\n\nAlternatively, here's another angle: gas thermometric speeds. At
Rollout 24, Act: -5.606
covering an edge in the dominance hypergraph.\n\nAnother thought inspired by linear algebra: Imagine creating a matrix where
Rollout 80, Act: -5.625
is there a part where I messed up units? Let's check units:\n\nEnergy per neutrino:
Rollout 32, Act: -5.630
there's a pattern or a recurrence relation here. Let me think. Maybe first compute a few powers of
Rollout 3, Act: -5.649
161.4 m/s.\n\nAs an assistant, I need to perhaps suggest both possibilities, but
Rollout 28, Act: -5.678
derived above: 4b2h2 + 4ab hk +4a2k2 -
Rollout 22, Act: -5.762
proceed in the absence of that info.\n\nWait another thought. You can also see that in textbooks, while
Rollout 24, Act: -5.792
size by using a property that is preserved.\n\nAnother thought: For each minimal saddle pair (R, C
Rollout 80, Act: -5.824
00 km = 1.2 × 10^8 cm.\n\nVolume = (4/
Rollout 35, Act: -5.824
19 and 25: not.\n\nTrial 3: n−17=-1, n
Rollout 46, Act: -5.851
2 / ��2 - ��2 / 2 (x2 + y2)\n\nThen:\n\n
Rollout 81, Act: -5.874
:\n\n20h / (51 - h^2) = 22/7 -> cross
Rollout 98, Act: -5.946
the answer choices only include options A to D. Let me confirm each system again:\n\n1. Rejected

UP_PROJ

Top 16 Positive Activations
Rollout 30, Act: 3.368
sum is sum_{n=1}^\infty [1/(a_{n+1}-1
Rollout 30, Act: 3.271
=1 }^\infty 1/(a_k(a_k -1 )) for the same reason as
Rollout 30, Act: 3.263
same as Sum_{k=1 }^\infty 1/(a_k(a_k -1 ))
Rollout 1, Act: 3.254
x1 - x2, 2y⟩ = ||x1||2 - ||x2||
Rollout 30, Act: 3.218
}^{\infty} \frac{b_n (b_{n+2} - b_n
Rollout 16, Act: 3.187
] = sum_{m=0}^\infty [1 - (1 - (1/2
Rollout 11, Act: 3.181
)( r e^{iθ}) ) = r2 + (1/r) e^{-iθ}
Rollout 69, Act: 3.174
sum S would telescope:\n\nS= �� α_i ( sec(theta_{i-1}) - sec
Rollout 66, Act: 3.152
n -12n *n )=gcd(n, n )=n.\n\nTherefore, unless n
Rollout 91, Act: 3.147
) can be expressed as |u × v| / |u ��� v |? Wait,
Rollout 96, Act: 3.147
098,815= -3,720,873\n\nHence
Rollout 30, Act: 3.139
1 + Sum_{k=1 }^\infty [1/(a_k -1 ) -
Rollout 36, Act: 3.121
-10,800 => -1,728\n\nSo:\n\n2s2 -
Rollout 1, Act: 3.115
x and x',\n\n||x - x'||2 = d2,\n\nand ⟨x, x'⟩
Rollout 25, Act: 3.102
:\n\nr_p = (h2 / mu) / (1 + e )\n\nAs h = r*v
Rollout 18, Act: 3.099
=7a +25 => 7a=4-25= -21 => a
Top 16 Negative Activations
Rollout 4, Act: -4.946
4)/2≈6 pF. Wait, but when we did the partial derivatives, we got ~
Rollout 54, Act: -4.953
weaker, that means HS− is a stronger base. Wait, wait. No: wait, the lower
Rollout 44, Act: -4.955
question is, is this the maximum possible work? But wait, when compressing irreversibly against a
Rollout 53, Act: -4.966
to double-check whether P is really the centroid. But the problem states that medians AD and CE intersect
Rollout 48, Act: -4.985
to get 0x907F? But this is the alternative method.\n\nWait, perhaps this
Rollout 95, Act: -5.017
, so maybe the ratio k relates to area. But according to the prior reasoning, the area of triangle
Rollout 39, Act: -5.050
321 ones followed by a 0. But in our case, wait, but 10
Rollout 44, Act: -5.066
be in an irreversible process compared to reversible processes? But according to the previous examples, yes. For example
Rollout 27, Act: -5.098
� 0.866 k T. But that's if E is (1/2)
Rollout 86, Act: -5.111
/4 + 0=pi/4.\n\nBut with the previous sign:\n\nThus returning, the integral
Rollout 44, Act: -5.151
a higher constant pressure, you have more work. But if compressing reversibly, which is a slow
Rollout 74, Act: -5.173
is the coefficient for the heeling error correction. But wait, maybe K_b and K_s here are
Rollout 23, Act: -5.220
reciprocal of the decay constant is the mean lifetime. But the half life relations are.\n\nBut let's double
Rollout 41, Act: -5.242
, if considered, numbers with exactly one digit.\n\nBut if we take all these summed up:\n\n1-digit
Rollout 23, Act: -5.389
1/(sum of 1/T_i). No, no. The decay constants sum.\n\nHang on,
Rollout 23, Act: -5.421
sum of 1/T_i). No, no. The decay constants sum.\n\nHang on, no.

DOWN_PROJ

Top 16 Positive Activations
Rollout 18, Act: 7.075
=7a +25 => 7a=4-25= -21 => a
Rollout 64, Act: 6.018
is integer, A=2*1 -3= -1. Then the general term is a_n
Rollout 75, Act: 5.990
12 ��3 mod8 → a �� -9 �� -1 ��7 mod
Rollout 35, Act: 5.964
0. Solutions: b = [-7 ± sqrt(49 -40)]/2 = [-
Rollout 12, Act: 5.887
) = K(k), with dk/dnu = ( -nu / sqrt(1 - nu^2
Rollout 75, Act: 5.869
a + 20 ��3→ a≡-17≡7 mod8. No.\n\n
Rollout 93, Act: 5.850
formula gives sqrt(3/3.14≈0.97 whereas actual formula gives33
Rollout 12, Act: 5.717
u) = K(k), with dk/dnu = ( -nu / sqrt(1 - nu^
Rollout 19, Act: 5.513
k=1... then n=1-5= -4 which isn’t positive. Disregard
Rollout 18, Act: 5.418
=7a +15 => 7a=8-15 = -7 => a=
Rollout 87, Act: 5.415
(1 + r2). Then du/dr = [ (1/(2 sqrt(1 + r2
Rollout 87, Act: 5.353
≈0.374, 1/r≈2.67\n\nSample 2: (
Rollout 49, Act: 5.341
, which is true in Hausdorff spaces since points are closed. But that's not helpful.\n\n
Rollout 49, Act: 5.334
know that in Hausdorff spaces, compact sets are closed. So every compact subset of X is
Rollout 64, Act: 5.317
2*1 -3 +1*2^n= -1 +2^n. For n=1
Rollout 75, Act: 5.310
16≡3 mod8→ a ��-13≡3 mod8. But a
Top 16 Negative Activations
Rollout 53, Act: -4.287
36=828? 23*30=690, 23
Rollout 46, Act: -4.299
written?\n\nWait, perhaps since the problem mentions expressing them "in terms of U", they just want the
Rollout 46, Act: -4.308
that. Hmm, but the problem says "express... in terms of U". If U is given as
Rollout 93, Act: -4.353
=0 to1007, 1008 terms. Apologies, getting tangled up
Rollout 46, Act: -4.432
t\n\nSimilarly for Acc_y.\n\nBut the problem says "in terms of U". If we take U
Rollout 25, Act: -4.438
.118)≈ 1.057\n\nHence denominator is 1+1
Rollout 46, Act: -4.442
would be the gradients.\n\nBut again, the question says "Express d2x/dt2 and d
Rollout 46, Act: -4.451
Coriolis), and the problem states to express them in terms of U. But unless the velocity terms
Rollout 46, Act: -4.469
of U and the velocities. However, the problem says "in terms of U", which is a bit
Rollout 72, Act: -4.478
even n: we can take 1,1,-1 and sum to1. But for odd n
Rollout 46, Act: -4.510
leave it at that. Hmm, but the problem says "express... in terms of U". If U
Rollout 30, Act: -4.529
2015*2014 =2015*(2000 +1
Rollout 46, Act: -4.632
terms. However, since the problem only mentions expressing them “in terms of U”, I'm now beginning
Rollout 82, Act: -4.921
. Therefore, the problem is equivalent to splitting the symbols into three groups, and for each group, creating
Rollout 82, Act: -4.975
best.\n\nSo the approach would be to partition the symbols into three groups such that when each group is Huffman
Rollout 82, Act: -5.958
's think again.\n\nSuppose we partition the six symbols into three groups: G1, G2,

Layer 51

GATE_PROJ

Top 16 Positive Activations
Rollout 73, Act: 6.674
0 = same expression.\n\nSimplify:\n\nLet's compute the coefficients:\n\n- (n+1)
Rollout 64, Act: 6.486
)/2 *2^n\n\nSimplifying, let's substitute n=2010:\n\na
Rollout 5, Act: 6.432
+ x)) = 125H\n\nLet me factor H out:\n\n(x + 17
Rollout 28, Act: 6.429
(2k -b)) b2\n\nLet me factor out B:\n\nB ab = B [ -
Rollout 28, Act: 6.426
/ (2k -b)) b2\n\nLet me factor out B:\n\nB ab = B [
Rollout 5, Act: 6.335
175)/(x + 75)\n\nLet me write the entire equation:\n\n(x + 1
Rollout 95, Act: 6.312
complicated. Maybe coordinate system is better.\n\nLet me place point A at the origin (0,0).
Rollout 11, Act: 6.267
i2θ}))(re^{iθ})\n\nLet me compute step by step:\n\n- a + b
Rollout 64, Act: 6.250
1 ) <=2^{2012}\n\nLet's substitute z= x -1, since x
Rollout 84, Act: 6.235
/9)*(27/32). Let's compute (2/9)*(27/3
Rollout 11, Act: 6.222
2θ}))(re^{iθ})\n\nLet me compute step by step:\n\n- a + b =
Rollout 73, Act: 6.218
- 0 = same expression.\n\nSimplify:\n\nLet's compute the coefficients:\n\n- (n+1
Rollout 17, Act: 6.193
5.29 *5.29: 5^2=25, 5
Rollout 95, Act: 6.179
�3 k /490 ∈ Q. Let me denote r = ��3k /4
Rollout 66, Act: 6.170
the length of this altitude C1C2 would be (product of the legs)/hypotenuse.\n\n
Rollout 5, Act: 6.163
75) = 125\n\nLet me let y = x + 75.
Top 16 Negative Activations
Rollout 37, Act: -4.000
divides $\det A$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I just came across
Rollout 62, Act: -4.010
is \(\boxed{132}\).<|im_end|>
Rollout 94, Act: -4.017
correct answer is \(\boxed{B}\).<|im_end|>
Rollout 2, Act: -4.028
is \(\boxed{109}\).<|im_end|>
Rollout 44, Act: -4.042
Final Answer: \(\boxed{ABC}\)<|im_end|>
Rollout 29, Act: -4.043
Final Answer**\n\boxed{A}\n\n<|im_start|>answer\nGiven a gallon milk jug with a plugged handle due
Rollout 57, Act: -4.045
. \n \n*A. Csaszar*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show
Rollout 91, Act: -4.047
|\) is \(\boxed{2}\).<|im_end|>
Rollout 11, Act: -4.065
\boxed{(-1, 3)}\n\n<|im_start|>answer\nGiven three complex numbers \(a, b,
Rollout 47, Act: -4.080
What is the value of $N$?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so there's this
Rollout 4, Act: -4.084
energy stored in this combination of capacitors?\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to
Rollout 28, Act: -4.132
a^2b^2 =0}\)<|im_end|>
Rollout 95, Act: -4.136
**Final Answer**\n\boxed{0}\n\n<|im_start|>answer\nTriangle \(ABC\) has side lengths \(AB
Rollout 27, Act: -4.151
component of velocity along the line of sight).\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I've got
Rollout 87, Act: -4.174
}$, and any necessary numerical constants.\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find
Rollout 48, Act: -4.217
is valid. \boxed{No}\n\n<|im_start|>answer\nTo determine if the IPv4 header contains an error

UP_PROJ

Top 16 Positive Activations
Rollout 70, Act: 5.077
-digit numbers, we need to check.\n\nBut this is getting complicated. Maybe instead of trying m=4
Rollout 49, Act: 5.064
a space would have other properties. Hmm. This seems a bit vague.\n\nWait, let me get back
Rollout 5, Act: 5.013
exceeding x2/100. \n\nFirst, let me recall some trapezoid properties.
Rollout 57, Act: 4.983
of open sets narrowing down on C. But this is vague. Alternatively, consider that in the density topology
Rollout 36, Act: 4.950
find the distance from P to A. \n\nFirst, maybe I should visualize the cube. Let's assign
Rollout 30, Act: 4.931
compute 2015 *2014:\n\nNote that 2014=2
Rollout 93, Act: 4.878
+1)) depending on n.\n\nWait, that's too intricate.\n\nAlternatively, another idea related to the
Rollout 66, Act: 4.770
Find $p$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through this problem
Rollout 59, Act: 4.753
392 share any common factors.\n\n123 factors: 3 * 41.\n\n3
Rollout 95, Act: 4.728
that have rational length.\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I need to solve
Rollout 10, Act: 4.716
2a + 3b, but that's not helpful. Wait, 0 is even,
Rollout 36, Act: 4.703
Find $AP.$\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to figure out how to
Rollout 95, Act: 4.689
segments P_kQ_k have rational lengths.\n\nFirst, let me visualize triangle ABC. Let me sketch it
Rollout 3, Act: 4.684
molecular speed of radon gas. Hmm, first, I remember there's a formula for the root mean
Rollout 93, Act: 4.674
/(k,1007). Wait, this line of thinking might not be too helpful.\n\nAlternatively
Rollout 15, Act: 4.654
can be such that all initial states die out. But maybe not?\n\nAlternatively, perhaps the good ns are
Top 16 Negative Activations
Rollout 17, Act: -4.376
.\n\nCalculating units:\n\n( T·m/A * J/T ) / m3 = (J·
Rollout 14, Act: -4.389
the integrals:\n\n- [I] = ��f dx ~ [f] * [x],\n
Rollout 32, Act: -4.397
78\n\nNow the coefficients of ��2:\n\n-670ω2 + (-67
Rollout 3, Act: -4.429
s2·mol). Then, dividing by M in kg/mol, the mol unit cancels, then
Rollout 89, Act: -4.464
of the children are girls. Let that parse. "2/5 of the children (girls)" rather
Rollout 14, Act: -4.515
] = ��f dx ~ [f] * [x],\n- [J] = ��
Rollout 85, Act: -4.527
can they be the same? Wait, the problem states "the set of numbers labeled on the points"
Rollout 41, Act: -4.529
9 = 81. But the problem says at most two different digits. Since a two-digit
Rollout 21, Act: -4.582
in S have image Z. Wait, no: "image exactly Z" means that the image is the
Rollout 23, Act: -4.587
keeping units in minutes.\n\nCompute activity in Bq (per second), so time t=10 minutes
Rollout 26, Act: -4.598
two decimal places. Wait the units in the problem says 'X kJ mol^−1', and
Rollout 46, Act: -4.643
of U and the velocities. However, the problem says "in terms of U", which is a bit
Rollout 69, Act: -4.736
the sum can be written as (1/4)(sec(theta_{i-1}) - sec(theta
Rollout 88, Act: -4.805
104. But the wording here is only "weights", which typically refers to the multiplicative parameters
Rollout 69, Act: -4.820
)Sum_{i=0}^\infty (sec(theta_{i-1}) - sec(theta
Rollout 69, Act: -4.903
S would telescope:\n\nS= �� α_i ( sec(theta_{i-1}) - sec(theta

DOWN_PROJ

Top 16 Positive Activations
Rollout 62, Act: 5.728
starting from \( m = 1 \) and moving upwards until we find the desired number makes sense.\n\n
Rollout 42, Act: 5.625
are treated with acid. Phosphoric acid is a strong acid, and in such conditions, alcoh
Rollout 42, Act: 5.544
ols are treated with acid. Phosphoric acid is a strong acid, and in such conditions, al
Rollout 81, Act: 5.352
0.\n\nCompute terms:\n\n484x2 -x(533*102 +
Rollout 42, Act: 5.342
ates the hydroxyl group, converting it into a good leaving group (water). Then, the elimination
Rollout 42, Act: 5.229
The hydroxyl group is protonated, forming a good leaving group (water). Upon departure of water
Rollout 30, Act: 5.208
15*(2000 +14)=2015*2000 +
Rollout 75, Act: 5.197
.n^2 - (n-1)^2=2n-1. But for odd squares,
Rollout 42, Act: 5.189
ocation. Then, the base (water acting as a weak base) would deprotonate a nearby
Rollout 42, Act: 5.140
propyl ring. When the acid is added, it protonates the hydroxyl group, converting it
Rollout 63, Act: 5.134
f(1, 1) = 1 + 0 + 0 = 1\n\
Rollout 44, Act: 5.108
abatic expansion, pressure drops more). Therefore, area under isothermal curve is larger; so work done
Rollout 42, Act: 5.006
in aqueous H3PO4, which is a good acid, it's more likely an E1
Rollout 75, Act: 4.931
m2 = (m-1)2+ (difference) which is even? So (m-
Rollout 42, Act: 4.926
lopropyl ring. When the acid is added, it protonates the hydroxyl group, converting
Rollout 75, Act: 4.922
as m2 = (m-1)2+ (difference) which is even? So (m
Top 16 Negative Activations
Rollout 2, Act: -4.251
impossible because the numbers are huge. So, perhaps there's a pattern or periodic behavior we can exploit when
Rollout 95, Act: -4.253
k/2450=5*sqrt(3)*k/2450.\n\nThus
Rollout 51, Act: -4.276
be removed by adjusting the sign appropriately.\n\nLet me recall that the absolute value of sin(x - ��
Rollout 16, Act: -4.308
But calculating this sum directly might be challenging. So perhaps we can use inclusion-exclusion or some combinator
Rollout 72, Act: -4.334
to 1. \n\nFirst, I need to recall that the roots of unity of order \( n \
Rollout 23, Act: -4.369
57)^7. Hmm, impractical. Perhaps use approximation.\n\nAlternatively, maybe use the natural logarith
Rollout 87, Act: -4.398
perhaps this would get too involved.\n\nAlternatively, perhaps try numerical integration for the dimensionless constant. Since the
Rollout 91, Act: -4.421
of triangle). Wait, not exactly. Let's recall that in triangle, cot B = adjacent/opposite
Rollout 21, Act: -4.437
x_i; you cannot quantify existentially. Unless there's a way to express surjectivity.\n\nWait
Rollout 11, Act: -4.516
) ) ]\n\nBut this is very complicated, perhaps there's no immediate simplification. However, if I
Rollout 97, Act: -4.534
total volume (concentration times RT?), but perhaps simpler to relate via the stoichiometry.\n\nAlternatively,
Rollout 40, Act: -4.577
Or perhaps using complex numbers? Hmm. Let me recall that rotations can be represented using complex numbers by multiplying
Rollout 16, Act: -4.621
ling, multiple conversions possible.\n\nWait, let's check a sample. What's the expectation until one sap
Rollout 93, Act: -4.639
My goal is to evaluate this sum. Let me think about how to approach it. \n\nFirst off,
Rollout 66, Act: -4.983
to find the perimeter p.\n\nAlright, let me recall that in a right-angled triangle, the altitude
Rollout 11, Act: -5.175
y )\n\nBut now this seems really complicated. Perhaps there's another substitution?\n\nAlternatively, setting variables in terms

Layer 52

GATE_PROJ

Top 16 Positive Activations
Rollout 39, Act: 2.076
+ 9999 + \cdots + \underbrace{99\ldots
Rollout 44, Act: 2.040
=\mathrm{T}_{2}$, and (ii) positive, if it is expanded reversibly under ad
Rollout 91, Act: 1.919
(\overrightarrow{CB} = (-2, 0)\).\n\n3. **Calculations**:\n
Rollout 17, Act: 1.896
\mathbf{I}=\int \mathbf{B}(\mathbf{r})|\
Rollout 12, Act: 1.713
x^2)}}\] \n(where the positive value of the square root is taken) is monotonically
Rollout 12, Act: 1.703
^2)}}\] \n(where the positive value of the square root is taken) is monotonically decreasing
Rollout 75, Act: 1.674
+ 9\) and \(9 = 4 + 4 + 1\).\n\n4. **
Rollout 45, Act: 1.598
{} \minus{}4$ and $ f(x\plus{}y) \equal{} f
Rollout 33, Act: 1.582
$ $\leq$ $k$ $\leq$ $5$ , at least one of the
Rollout 12, Act: 1.576
^2x^2)}}\] \n(where the positive value of the square root is taken) is mon
Rollout 44, Act: 1.508
, and (ii) positive, if it is expanded reversibly under adiabatic conditions with $\
Rollout 30, Act: 1.494
\right)$ and $b_{n+1}=a_{n} b_{n-1}-
Rollout 2, Act: 1.468
999 \times \cdots \times \underbrace{99\cdots9
Rollout 28, Act: 1.440
the conic equation, \(B^2 - 4AC\), is calculated to be positive for
Rollout 2, Act: 1.435
times \cdots \times \underbrace{99\cdots9}_{\text{9
Rollout 44, Act: 1.407
work done on the gas is maximum when it is compressed irreversibly from (p2, V2
Top 16 Negative Activations
Rollout 32, Act: -7.298
there's a pattern or a recurrence relation here. Let me think. Maybe first compute a few powers of
Rollout 46, Act: -7.315
or some combination, but not sure.\n\nWait, let's go back. The problem says:\n\n"Express
Rollout 55, Act: -7.325
n, 0 when raised to the power. Let me check that. If I think about A squared
Rollout 58, Act: -7.344
seems complicated, but perhaps we can simplify it. Let's note that P(0) is the product
Rollout 23, Act: -7.366
, maybe the user didn't specify units conversion. Let me check again thoroughly.\n\nLet's recast all
Rollout 67, Act: -7.393
finding the general quadratic term, then find terms. Let's proceed both ways.\n\nFirst method: assume the
Rollout 98, Act: -7.404
the answer choices only include options A to D. Let me confirm each system again:\n\n1. Rejected
Rollout 95, Act: -7.430
, seems complicated. Maybe coordinate system is better.\n\nLet me place point A at the origin (0,
Rollout 87, Act: -7.439
] ds dt * a\n\nWait, no. Let's check again:\n\nOriginal expression:\n\nStarting from V
Rollout 46, Act: -7.447
U/∂x, maybe with a sign. Let me check. If U is the potential energy,
Rollout 66, Act: -7.449
. Each altitude length is multiplied by a factor. Let's check.\n\nFirst altitude (from C0):
Rollout 9, Act: -7.454
I'm confused. Wait squares have four sides. Let me actually assign coordinates to the square to make this
Rollout 73, Act: -7.474
0. Then, perhaps compute the first sum. Let me compute that.\n\nLet me denote S1 =
Rollout 30, Act: -7.490
That telescopes if we write it as such. Let's check:\n\nSum from n=1 to �
Rollout 43, Act: -7.503
where maybe there's a different arrangement.\n\nWait, let me check again the answer options as given:\n\nAnswer
Rollout 30, Act: -7.979
led us to a more handle-able form.\n\nLet me recap:\n\nSo, the original term in the

UP_PROJ

Top 16 Positive Activations
Rollout 52, Act: 6.919
: (-501+505)/2=4/2=2. So sum is
Rollout 52, Act: 6.530
2 = (1007 -3)/2=1004/2=50
Rollout 52, Act: 6.464
(4028 -1 +1)/2 =4028/2=20
Rollout 68, Act: 6.428
x2= (9 -4 -3)/2=2/2=1. So both x
Rollout 52, Act: 6.363
)/2= (76 -52)/2=24/2=12. So
Rollout 99, Act: 6.324
.14794)/0.25=0.05103/0
Rollout 52, Act: 6.311
2= (212 -18)/2=194/2=97.
Rollout 81, Act: 6.273
=( -70 +172)/22=102/22=51
Rollout 45, Act: 6.265
-4= -64 +64 -4= -4 ✔️. Okay, matches.\n\n
Rollout 68, Act: 6.226
[-(4 - 8)] / 2 = [4] /2 = 2. So
Rollout 52, Act: 6.219
028/1 -1 +1)/2 = (4028 -1 +1
Rollout 99, Act: 6.126
2.19897)/0.5≈-0.19897/
Rollout 82, Act: 6.013
10 +16 +16)/26 = 42/26 ��1
Rollout 75, Act: 5.967
so (1 +1 +1-3)/4= (3-3)/4=0mod
Rollout 99, Act: 5.914
19897)/0.5 �� (-0.19897)/0
Rollout 0, Act: 5.914
2,3,5,7 are even?\n\nWait floor operations:\n\nFor prime 2:
Top 16 Negative Activations
Rollout 30, Act: -4.692
that the sequences are defined recursively with overlapping terms, maybe there's a pattern here or a way to express
Rollout 93, Act: -4.703
, and we're summing over many terms, maybe we can approximate each term using Laplace's method
Rollout 82, Act: -4.708
? That's the crux. Let me think if there's a systematic way.\n\nOne heuristic is to
Rollout 14, Act: -4.721
terms of $J$ and $K$.\n\nMaybe Cauchy-Schwarz on $I$
Rollout 40, Act: -4.760
origin, and then the translation is a shift, maybe we can represent this as a combination of rotation and
Rollout 69, Act: -4.835
i increases, both angles go to zero. So perhaps we can approximate terms for large i? However,
Rollout 69, Act: -4.851
the angles involve divisions by 3^i, perhaps the sum has some recursive relation. For example,
Rollout 34, Act: -4.932
convolution kernel because it depends on t - s. Maybe I can convert these integral equations into differential equations using
Rollout 33, Act: -4.964
angements but generalized to different positions.\n\nAlternatively, maybe I can model this problem using inclusion-exclusion,
Rollout 75, Act: -5.007
easier. However, since six squares is allowed, perhaps using Lagrange's theorem as a basis, then
Rollout 69, Act: -5.013
(3 theta_i) ]\n\nWe have to see if each term can be rewritten in terms of a telesc
Rollout 69, Act: -5.022
)/cos(3 theta_i). Let's see if this can be manipulated.\n\nPerhaps use the identity for
Rollout 69, Act: -5.047
need an exact value.\n\nAlternatively, let's see if we can rewrite the term in terms of tangent or
Rollout 77, Act: -5.094
third property, which seems like a functional equation. Maybe I can use this to reduce f(14
Rollout 63, Act: -5.164
\( f(p, q) \). Alternatively, perhaps express it in terms of \( p \) and
Rollout 83, Act: -5.273
seems like a linear recurrence relation. Let me see if there's a pattern here. If we have the

DOWN_PROJ

Top 16 Positive Activations
Rollout 90, Act: 6.503
.\n\nFirst, compute how much 2018 divides into10,080,0
Rollout 20, Act: 6.331
46 divides 966? 46 divides into 966? 46
Rollout 62, Act: 6.116
: 7*2075=14525, 14531
Rollout 16, Act: 6.105
divided by 217. 217*5=1085, so
Rollout 70, Act: 6.005
: 21.857. 7*21=147, remainder6
Rollout 96, Act: 5.944
7 by2197:\n\n2197*31=2197*3
Rollout 87, Act: 5.937
(r,θ) [ (1/r) * r2 sinθ dr dθ dφ ]\n\nBut
Rollout 95, Act: 5.759
??? Wait.\n\nWait, hold on. Let me recast this.\n\nPoint Pk is along AB.
Rollout 71, Act: 5.704
57... which is not an integer. 11? 2011 ��1
Rollout 90, Act: 5.684
let's compute how many times 2018 goes into 336,000
Rollout 62, Act: 5.589
's test divide by 17: 17*59=1003?
Rollout 20, Act: 5.525
6 divides into 966? 46 *21 = 966 (4
Rollout 65, Act: 5.517
2 if and only if a2 has in its prime factors 3,5,7,19
Rollout 98, Act: 5.479
have a shorter main-sequence time. Let me recalculate: main-sequence lifetime formula: t ~
Rollout 99, Act: 5.469
would give x4. But if the user is bound to do two iterations, would the answer be x
Rollout 37, Act: 5.469
so when reducing modulo p, the matrix is over the field F_p. Now, the problem reduces to
Top 16 Negative Activations
Rollout 78, Act: -5.699
that is 110/20 = 11/2 = 5.5 minutes
Rollout 3, Act: -5.717
0\n\nsqrt(27,950)= approximately 167.2 m/s.
Rollout 56, Act: -5.821
1√3 /12)^2 = (121*3)/144 =
Rollout 3, Act: -5.834
225,753 ) �� 475 m/s. Which is correct.\n\n
Rollout 81, Act: -5.904
9584) = let's check: 172^2=2958
Rollout 3, Act: -5.963
Let's see.\n\n168^2 = (170 - 2)^2 =
Rollout 23, Act: -5.970
-life (since the beta decay half-life is only 0.098 min, which is the
Rollout 78, Act: -6.018
60 = 22/40 = 11/20 minutes.\n\nSimilarly, the
Rollout 3, Act: -6.026
(28408.12)= 168.546... Hence,
Rollout 3, Act: -6.100
28,408.12)= 168.5\n\nThus, answer is
Rollout 56, Act: -6.211
] = sqrt[49/36 + (121*3)/144 ]
Rollout 69, Act: -6.254
1938, squared �� 0.000376. cos(3
Rollout 56, Act: -6.289
11√3 /12)^2 = (121*3)/144 =
Rollout 3, Act: -6.536
Then sqrt(225,298)= approx 474.6 m/s.
Rollout 69, Act: -6.800
.01938, squared �� 0.000376. cos
Rollout 3, Act: -6.943
(225,298)= approx 474.6 m/s. So about

Layer 53

GATE_PROJ

Top 16 Positive Activations
Rollout 8, Act: 7.357
*9. 31 and 126 are coprime, so yeah, simplified.\n\n
Rollout 61, Act: 7.338
3 and 780 are coprime. Let's check.\n\nDivisors of 13
Rollout 47, Act: 7.328
found v_air = (4/3) vt. Let's substitute:\n\np0 - p = (
Rollout 58, Act: 7.247
018)(2019)/2. Calculating that: 2018 *
Rollout 72, Act: 7.233
1)/4 * 2? Let me verify:\n\nWe know that cos(36°)
Rollout 16, Act: 7.223
ators are 217 and 63. Factoring:\n\n217 = 7*
Rollout 16, Act: 7.195
7880 and 1953 are coprime as established earlier, so the
Rollout 20, Act: 7.178
7*23*43*47. To check:\n\nThe LCM takes the max ex
Rollout 20, Act: 7.176
common multiple of 42 and 46. Let me compute that.\n\n42 factors into
Rollout 71, Act: 7.165
92+492 +102. Let me compute that:\n\n392=
Rollout 9, Act: 7.148
\n\nSo 20y = 100 => y = 5.\n\nTherefore, the y
Rollout 11, Act: 7.131
/(ab)) + (1/(ab))/a. Simplify each term: a/b is the same
Rollout 38, Act: 7.092
. Wait, 196/672= (28*7)/(28*
Rollout 77, Act: 7.082
(52/38)*266. Calculate that: 52/38 simpl
Rollout 17, Act: 7.078
.29 ×10^-11 m. Therefore, a03=(5.29
Rollout 16, Act: 7.077
7880/1953.\n\nLet me check if 7880 and
Top 16 Negative Activations
Rollout 25, Act: -1.494
649e7)^2 = 7.479e15\n\nThen numerator:
Rollout 33, Act: -1.532
4, 5, 6$ such that for each $k$ with $1$ $\le
Rollout 48, Act: -1.547
treating them as unsigned 16-bit: 46,610. Adding 17
Rollout 48, Act: -1.549
value) in the IP header: 4500 0034 B612
Rollout 48, Act: -1.567
64).\n\n0x5BC6 is 23444 (approx: 5*
Rollout 23, Act: -1.602
0.098} \approx 7.0724 \, \text{min
Rollout 80, Act: -1.667
10^24 �� 7.238 × 10^24
Rollout 17, Act: -1.711
5.29*5.29=27.9841, then 2
Rollout 38, Act: -1.741
5/124 x≈1.25x.\n\nSo Length L≈1.25
Rollout 38, Act: -1.764
155/124≈1.25. So, x<=1.25
Rollout 72, Act: -1.788
cos(72°) �� 0.618, so 1 + 0
Rollout 44, Act: -1.854
\n\n[D] The change in internal energy of the gas is (i) zero, if it is expanded
Rollout 86, Act: -1.880
5708)^5 /5 �� 8.89 /5 �� 1
Rollout 38, Act: -2.103
55/124 x≈1.25x.\n\nSo Length L≈1.2
Rollout 78, Act: -2.124
along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east
Rollout 95, Act: -2.213
}, P_{1}, P_{2}, ... , P_{2450} = B$

UP_PROJ

Top 16 Positive Activations
Rollout 25, Act: 4.466
hyperbolic trajectory:\n\ne = sqrt(1 + (2 ε h2)/mu2) )\n\n
Rollout 22, Act: 4.400
6.02214076 = approximately 0.3347, then
Rollout 25, Act: 4.195
/ μ) / (1 + sqrt(1 + (2 * ε * h2)/μ2
Rollout 27, Act: 4.146
gas, T_c is given by:\n\nT_c = ( (n^(2/3) h^
Rollout 25, Act: 4.087
/2\n\nThen, e = sqrt(1 + 2 epsilon h2 /mu2 )\n\nYes
Rollout 25, Act: 3.996
/ μ) / (1 + sqrt(1 + 2*ε*h2/μ2))\n\n
Rollout 25, Act: 3.869
* 1 / [1 + sqrt(1 + (2 epsilon h2)/mu2 +
Rollout 27, Act: 3.868
another approach might be to use that k T_c = (1/2) m v_rms^
Rollout 25, Act: 3.778
trajectory, the perigee is:\n\nr_p = (h2 / μ) / (1 +
Rollout 25, Act: 3.681
2/2\n\nThen, e = sqrt(1 + 2 epsilon h2 /mu2 )\n\n
Rollout 25, Act: 3.658
.\n\nThe specific orbital energy ε is given by ε = v2 / 2 - μ / r.\n\n
Rollout 27, Act: 3.634
_rms = 5.54 × 10^-17 J.\n\nWe need to find
Rollout 25, Act: 3.633
is positive, ε = v2 / 2 - μ / (2a), where a is the
Rollout 95, Act: 3.629
2 + (Δy)^2 ) = sqrt( (4/9k2)/(245
Rollout 25, Act: 3.628
formula for an orbit's closest approach:\n\nr_p = (h2 / μ) / (1 +
Rollout 22, Act: 3.628
6.02214076 = approx 0.000334
Top 16 Negative Activations
Rollout 9, Act: -6.562
.\n\n**Final Answer**\n\boxed{25}\n\n<|im_start|>answer\nSquare \(AIME\) has sides
Rollout 19, Act: -6.568
}\n\n**Final Answer**\n\boxed{78}\n\n<|im_start|>answer\nTo solve the problem, we need
Rollout 2, Act: -6.574
**Final Answer**\n\boxed{109}\n\n<|im_start|>answer\nTo find the remainder when \(9
Rollout 13, Act: -6.591
**\n\boxed{187.12}\n\n<|im_start|>answer\nTo determine how much more money Jo
Rollout 11, Act: -6.599
Answer**\n\boxed{(-1, 3)}\n\n<|im_start|>answer\nGiven three complex numbers \(a,
Rollout 48, Act: -6.607
. The checksum is valid. \boxed{No}\n\n<|im_start|>answer\nTo determine if the IPv4 header
Rollout 66, Act: -6.613
**Final Answer**\n\boxed{182}\n\n<|im_start|>answer\nGiven a right triangle \(ABC_
Rollout 81, Act: -6.619
{\dfrac{510}{11}}\n\n<|im_start|>answer\nGiven triangle \(ABC\) with \
Rollout 36, Act: -6.645
**Final Answer**\n\boxed{192}\n\n<|im_start|>answer\nGiven a cube with edges \(\
Rollout 91, Act: -6.652
}.\n\n**Final Answer**\n\boxed{2}\n\n<|im_start|>answer\nGiven triangle \(ABC\) with median
Rollout 95, Act: -6.722
}。\n\n**Final Answer**\n\boxed{0}\n\n<|im_start|>answer\nTriangle \(ABC\) has side lengths
Rollout 43, Act: -6.793
be \boxed{C}.\n\n\n**Final Answer**\n\boxed{C}\n\n<|im_start|>answer\nThe problem
Rollout 43, Act: -6.817
+ gamma |1>) \tensorproduct |0>\nAnswer Choices:\n(A) U_{C-NOT
Rollout 65, Act: -6.871
\n\n**Final Answer**\n\boxed{40}\n\n<|im_start|>answer\nGiven a right rectangular prism \( P
Rollout 97, Act: -6.954
**Final Answer**\n\boxed{230}\n\n<|im_start|>answer\nGiven the decomposition reaction \(2 \
Rollout 21, Act: -7.162
X}\n\n(b) \boxed{\text{Yes}}\n\n<|im_start|>answer\n(a) To show that Carl can

DOWN_PROJ

Top 16 Positive Activations
Rollout 2, Act: 7.058
109 mod125.\n\nSo summary so far:\n\nP≡5 mod8\n\nand P
Rollout 38, Act: 6.258
/5) sqrt(x2 + W2)\n\nMultiply both sides by 5:\n\n25(W +
Rollout 18, Act: 6.139
5. Since x is positive, this means x is a positive divisor of 35. The div
Rollout 94, Act: 5.996
**Final Answer**\n\boxed{B}\n\n<|im_start|>answer\nTo determine which states are entangled, we
Rollout 32, Act: 5.934
�2, where �� and ��2 are complex (primitive cube roots of unity). So �
Rollout 92, Act: 5.813
.78} diopters.\n\n<|im_start|>answer\nThe student is farsighted and uses glasses to
Rollout 34, Act: 5.751
) + exp(-2t) + C\n\nMultiply both sides by exp(t):\n\ny(t) =
Rollout 89, Act: 5.729
we can solve for T.\n\nSo from here, multiplying both sides by 5/3 to isolate T
Rollout 18, Act: 5.662
so 3y ��1 mod7. Multiply both sides by the inverse of 3 modulo7
Rollout 86, Act: 5.592
dfrac{\pi}{16}}.\n\n<|im_start|>answer\nTo find the minimum value of the integral \
Rollout 33, Act: 5.578
=3 contributes -48. The cumulative total so far is 720 (total) -
Rollout 89, Act: 5.568
/5)(G + 100)\n\nMultiply both sides by 5:\n\n5G =
Rollout 38, Act: 5.528
( x2 + W2 )/5\n\nMultiply both sides by 5*6.2 to
Rollout 25, Act: 5.519
a test particle when it's low mass, but as it becomes more massive, the two-body problem becomes
Rollout 87, Act: 5.517
For example, spherical coordinates. But since we are in a cube, spherical coordinates won’t align nicely.
Rollout 20, Act: 5.474
to understand the modulus. 2021 divided by, say, 43 is 2
Top 16 Negative Activations
Rollout 96, Act: -5.568
(5 -10.296,6 -(-6.710))=(-5
Rollout 16, Act: -5.571
(6,3)=20, (-1)^{4}=1, denominator 1 -1
Rollout 91, Act: -5.583
)\n\nVector BA: from B to A: (t - (-1), t - 0) =
Rollout 62, Act: -5.680
by13: 13* 1290.88=13*1
Rollout 95, Act: -5.748
ABC. Therefore, the ratio of areas would be proportional to the square of scaling factor.\n\nLet me note
Rollout 48, Act: -5.790
x907F? But this is the alternative method.\n\nWait, perhaps this is easier:\n\nWhen
Rollout 3, Act: -5.809
8,436). Let's see, 168^2 = 28,
Rollout 59, Act: -5.861
2 divided by 41? 41*9 = 369, 41
Rollout 96, Act: -5.868
9\n\n5t -12s =27 +1134/169\n\n
Rollout 39, Act: -6.076
0 +10^2 +... +10^{321} is equal to (1
Rollout 39, Act: -6.092
=(10 +10^2 +10^3)/9 - (3). Wait,
Rollout 16, Act: -6.126
3: C(6,3)*(-1)^4*[1/(1 - 1/8
Rollout 3, Act: -6.156
. At STP, the molar volume is 22.4 liters, which is for ideal
Rollout 68, Act: -6.244
4.\n\nThen quadratic equation becomes x2 + (4 - (-4))x+ 4= x
Rollout 96, Act: -6.422
69\n\nY_p - O_y =27 - (-1134/169
Rollout 26, Act: -6.948
elements in their standard states under standard conditions (usually 25°C and 1 atm).\n\nThe reaction

Layer 54

GATE_PROJ

Top 16 Positive Activations
Rollout 4, Act: 8.324
0.02 / 5.00 = 0.004.\n\nNow, since
Rollout 59, Act: 8.167
10/28 = 5/14.\n\nWait, hold on: 3+3
Rollout 72, Act: 8.141
°) is (sqrt(5)-1)/4? Wait, cos(72°) =
Rollout 31, Act: 7.851
√5)/2=(1 -√5)/2.\n\nThirdly, x=(1 -√5
Rollout 4, Act: 7.765
327 / 1200 �� 0.003606.
Rollout 95, Act: 7.750
= (21 + 5√3)/2. Hmm, seems complicated. Maybe coordinate system is
Rollout 8, Act: 7.749
31 +126=157.\n\nBut I want to cross-verify.\n\nWait
Rollout 72, Act: 7.714
(sqrt(5) - 1)/4 * 2. Wait, more accurately, cos(
Rollout 76, Act: 7.711
Hence, the answer is 929.\n\n**Final Answer**\n\boxed{929
Rollout 95, Act: 7.704
1) = (3√671)/2.\n\nSo the area of triangle ABC is 3
Rollout 96, Act: 7.694
5.562≈1.562\n\nSimilarly, y'=27 -63
Rollout 92, Act: 7.686
17200 / 4551 �� 3.77938
Rollout 58, Act: 7.679
= 2,019,000\n\n9 * 2019 =
Rollout 59, Act: 7.664
)*(3/7) = 3/28.\n\nFor k=1: There are three possibilities
Rollout 84, Act: 7.645
= 5 + 32 = 37. Wait, let me recheck the calculations.\n\n
Rollout 99, Act: 7.644
0206)/0.75 �� -0.8027467
Top 16 Negative Activations
Rollout 25, Act: -1.057
rigid body, the Roche limit is approximately d = 1.26 * R_earth *
Rollout 44, Act: -1.100
, but if it's compressed from V2 to V1, then ΔV = V1 - V
Rollout 25, Act: -1.101
body, the Roche limit is approximately d = 1.26 * R_earth * (
Rollout 83, Act: -1.136
obtained from the other by rotating the nonagon in the plane. Find the number of distinguishable acceptable
Rollout 17, Act: -1.142
_0^3} dV = \frac{2\mu_0}{3} \math
Rollout 65, Act: -1.159
$P_{}$ cuts $P_{}$ into two prisms, one of which is similar
Rollout 49, Act: -1.181
subspaces, so X is a continuous image of a certain space.\n\nThis seems too vague.\n\nAlternatively,
Rollout 25, Act: -1.219
67e-11\n\n= 6.13e14 /6.67
Rollout 99, Act: -1.222
.71145)≈-0.2192 -0.711
Rollout 49, Act: -1.288
, the space is countable or a quotient of a countable space, but no.\n\nAlternatively, consider
Rollout 70, Act: -1.306
.4286) �� 0.3567\n\nSo, (-1.
Rollout 49, Act: -1.404
Topology by Engelking, Theorem 3.8.2 on page 194:
Rollout 23, Act: -1.464
)+31*ln(10)=0.6419 +31*2.
Rollout 44, Act: -1.530
{p}_{1}, \mathrm{~V}_{1}\right)$ against constant pressure $\mathrm{
Rollout 99, Act: -1.564
.5)/ln(10)= (-0.6931471805
Rollout 49, Act: -1.598
Theorem 3.8.2 on page 194: Every regular Lindelöf

UP_PROJ

Top 16 Positive Activations
Rollout 28, Act: 6.738
2k -b)) b2\n\nLet me factor out B:\n\nB ab = B [ -2
Rollout 68, Act: 6.600
+ 4x + 4. Rearranging terms, we get:\n\nx^2 + (
Rollout 5, Act: 6.183
)) = 125H\n\nLet me factor H out:\n\n(x + 175)
Rollout 96, Act: 6.161
x - 120\nBring all terms to one side:\n5x - 12y
Rollout 7, Act: 6.159
18N =56N/3\n\nFactor x:\n\nx(A -54) +1
Rollout 5, Act: 6.058
5 + x) = 125\n\nFactor the second term:\n\nLeft side = (x +
Rollout 45, Act: 6.042
+ 4b + 4c\n\nSimplify left minus right:\n\n(30 - 4
Rollout 96, Act: 6.040
12y)/5 +5y =90\n\nMultiply all terms by 5 to eliminate denominator
Rollout 84, Act: 6.014
4 - 4x\n\nBring all terms to the left side or subtract 4 -4x
Rollout 11, Act: 6.004
r3 +1/r3 )=0\n\nFactor out 2i sinθ:\n\n2i sin
Rollout 11, Act: 6.003
2) sinθ cosθ = 0\n\nFactor out sinθ:\n\nsinθ [3 + (
Rollout 28, Act: 5.999
2h -a) + B k = 0 ... (1)\n\nC(2k - b
Rollout 25, Act: 5.997
μ_initial / r_p_initial = 0\n\nFactor out x:\n\nx [ v^4 - (
Rollout 31, Act: 5.983
:\n\n2 - x^2 - y = 0\n\nand\n\n- (1 + x y)
Rollout 5, Act: 5.978
75 + x)\n\nExpand the left-hand side:\n\nx^2 - 50x +
Rollout 77, Act: 5.969
(x, x + y)\n\nSo if I rearrange this, maybe I can express f(x, x
Top 16 Negative Activations
Rollout 9, Act: -3.969
h}x + 10 - \frac{50}{h}\).\n\n5. **Intersection
Rollout 96, Act: -3.974
197} = 31\n\]\n\nThus, the answer is:\n\[\n\boxed
Rollout 84, Act: -3.998
m = 5 \) and \( n = 32 \). Therefore, \( m + n
Rollout 96, Act: -3.998
= \frac{4106 \cdot 5 + 5697 \cdot
Rollout 66, Act: -4.036
the perimeter \(p = 182\).\n\nThus, the perimeter \(p\) is \(\
Rollout 0, Act: -4.051
6}{2} = 128\).\n\nThus, the number of rational numbers between 0
Rollout 66, Act: -4.058
to get the primitive triple \((84, 13, 85)\).\n\n4.
Rollout 95, Act: -4.065
3}k}{490}\) rational.\n\nThus, the number of segments \(\overline
Rollout 29, Act: -4.071
pressure at the bottom being \( \rho gh \).\n\n\[\n\boxed{A}\n\]<|im_end|>
Rollout 36, Act: -4.081
+ \left( \frac{s^2 + 18864}{2s} \
Rollout 64, Act: -4.134
2012} - k(2^{2009} - 1)\right)
Rollout 36, Act: -4.141
+ \left( \frac{s^2 + 8064}{2s} \right
Rollout 58, Act: -4.263
2037171\n\]\n\nThus, the total number of questions on the exam
Rollout 47, Act: -4.620
\implies N = 9\n \]\n\nThus, the value of \( N \) is
Rollout 8, Act: -4.762
m = 31\) and \(n = 126\). Thus, \(m +
Rollout 59, Act: -4.770
8}\n \]\n\n- For \(k = 3\) (all heads):\n \[\n

DOWN_PROJ

Top 16 Positive Activations
Rollout 42, Act: 8.192
since we have a good leaving group after protonation (water), perhaps E1 is possible. However,
Rollout 16, Act: 7.724
,953.\n\nTherefore, box that.\n\n**Final Answer**\n\boxed{7899
Rollout 3, Act: 7.285
� 28,436. Then take square root.\n\nsqrt(28,43
Rollout 54, Act: 7.247
an-1-olate, is a weaker base than hydroxide? No, no. Wait:
Rollout 98, Act: 7.196
become a white dwarf, it needs to go through the giant phases. But a star with 0.
Rollout 69, Act: 7.147
: 30° /9, that's 10/3°, and so on.
Rollout 54, Act: 6.950
8, so propionate is a much weaker base than hydroxide. So that correlates: hydrox
Rollout 95, Act: 6.892
⇒ x = 150/18 = 25/3 �� 8.
Rollout 11, Act: 6.832
, then A + B + C = 2πik for some integer k. But since a,
Rollout 42, Act: 6.832
ates the hydroxyl group, converting it into a good leaving group (water). Then, the elimination
Rollout 42, Act: 6.805
group is protonated, forming a good leaving group (water). Upon departure of water, a **ter
Rollout 54, Act: 6.793
xide is smaller and may engage in hydrogen bonding with solvent more. But maybe the lower basicity allows
Rollout 42, Act: 6.729
yl group, converting it into a good leaving group (water). Then, the elimination would occur, removing
Rollout 11, Act: 6.714
-real c as well.\n\nAlternatively, take a = re^{iθ}, b = re^{iθ
Rollout 89, Act: 6.710
improper, is the answer.\n\nThus, answering.\n\n**Final Answer**\n\boxed{\dfrac{2
Rollout 4, Act: 6.641
.00007696\n\nTake the square root: sqrt(0.00
Top 16 Negative Activations
Rollout 49, Act: -6.371
it σ-compact? No, because it's not second-countable and a σ-compact space which
Rollout 37, Act: -6.382
5^2 = 9*125 - 27*25 = 11
Rollout 5, Act: -6.419
base. The segment that joins the midpoints of the legs divides the trapezoid into two regions
Rollout 48, Act: -6.432
Just a identifier for fragmented packets.\n\nThen come Flags and Fragment Offset: "4000". Let
Rollout 27, Act: -6.449
26 +10) �� 3.318e-16\n\nThen divide
Rollout 70, Act: -6.474
in base-7: Since 7^3=343, 363-3
Rollout 44, Act: -6.502
to match the internal pressure at every step, hence maximum work is done. Wait, no: wait,
Rollout 70, Act: -6.544
(>333), so 7^3=343 (>333, so
Rollout 70, Act: -6.693
7 of 210:\n\n7^3=343>210. So
Rollout 19, Act: -6.744
about n=3. Then sum cubes1+8+27=36. Divided by
Rollout 98, Act: -7.021
)^3 �� 10^10 /27 �� 370 million years
Rollout 44, Act: -7.185
expansion into vacuum, no external pressure. Therefore, no work done (W=0). Also, since
Rollout 70, Act: -7.919
0 /7^3=2400/343≈6.997,
Rollout 37, Act: -8.423
^2 = 9*125 - 27*25 = 112
Rollout 37, Act: -8.486
=2, k2=3: 4*27 -8*9=108
Rollout 70, Act: -8.866
10 +d0 =2*(d3*343 +d2*49 +

Layer 55

GATE_PROJ

Top 16 Positive Activations
Rollout 22, Act: 3.456
/s. But in several decimal places.\n\n\n**Final Answer**\nThe mean molecular speed of hydrogen gas (H
Rollout 4, Act: 3.359
is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of the energy
Rollout 26, Act: 3.234
:\n\[\n\boxed{90.4}\n\]<|im_end|>
Rollout 4, Act: 3.204
[\n\boxed{0.88\%}\n\]<|im_end|>
Rollout 22, Act: 3.204
\text{m/s} \)\n\n**Final Answer**:\n\[\n\boxed{169
Rollout 23, Act: 3.183
\, \text{min}^{-1}\n \]\n - Beta decay half-life \( T
Rollout 42, Act: 3.173
group is protonated, forming a good leaving group (water). Upon departure of water, a **ter
Rollout 3, Act: 3.163
69} \, \text{m/s}\n\]<|im_end|>
Rollout 91, Act: 3.160
(t - 1, t)\) and \(\overrightarrow{CB} = (-2,
Rollout 27, Act: 3.155
\text{K}\n \]\n\n**Final Answer:**\nThe rms speed of the Be nuclei is
Rollout 15, Act: 3.132
of lamps \( n \) is a power of 2. This result is derived from properties of linear
Rollout 91, Act: 3.081
to find the value of \(|\cot B - \cot C|\).\n\n1. **Coordinate Setup**
Rollout 33, Act: 3.077
with \(1 \leq k \leq 5\), at least one of the first \(
Rollout 99, Act: 3.075
approximately2.1573.\n\n**Final Answer**\nThe decimal value of \( x_3 \
Rollout 23, Act: 3.065
)}{0.098} \approx 7.0724 \, \text
Rollout 92, Act: 3.050
3.78 diopters.\n\n**Final Answer**\nThe refractive power of the glasses lenses is
Top 16 Negative Activations
Rollout 51, Act: -8.126
1/7 ]\n\nWhich is:\n\n[1/9 -1/7] - [ -1/
Rollout 93, Act: -8.141
not an integral. Hmmm. \n\nWait, another thought: when dealing with sums over roots of unity
Rollout 30, Act: -8.145
a_{n+1}(a_{n+1} -1) b_n )\n\nTherefore,
Rollout 30, Act: -8.165
a_{n+1}(a_{n+1} -1 ) ]\n\nHmm. So now,
Rollout 44, Act: -8.166
work done on the gas is p1*(V2 - V1) if the external pressure is p
Rollout 12, Act: -8.171
) / sqrt( ( (t/ν)^2 -1 )(1 - t^2) )\n\n
Rollout 51, Act: -8.183
7\n\nSo the integral is:\n\n(1/9 -1/7) - ( -1/
Rollout 30, Act: -8.195
^∞ b_n (1/a_{n+1} - 1/a_{n+3}}
Rollout 12, Act: -8.240
( (1/ν) sqrt( (t2 - ν2)(1 - t2) ))
Rollout 51, Act: -8.241
the second bracket from the first:\n\n(1/9 -1/7) - (1/9
Rollout 30, Act: -8.244
term is b_n (1/a_{n+1} - 1/a_{n+3})
Rollout 51, Act: -8.346
+ 1/7 ]\n\n= (1/9 - 1/7) - (-1/
Rollout 12, Act: -8.360
(t2 - ν2) becomes (ν2 - ν2) =0, so that first
Rollout 30, Act: -8.420
a_{n+1}(a_{n+1} -1 ) = (a_{n-
Rollout 30, Act: -8.458
a_{n+1}(a_{n+1} -1) b_n ) = [ (
Rollout 51, Act: -8.815
/7\n\nTherefore:\n\nResult = (1/9 - 1/7) - (-1/

UP_PROJ

Top 16 Positive Activations
Rollout 20, Act: 4.019
}\) for all positive integers \( a \).\n\nFor \(\sigma(a^n) \equiv 1
Rollout 32, Act: 3.867
+ 0 + 1 = 2\)\n\nTotal sum for one cycle: \(2(1
Rollout 35, Act: 3.806
4)R(n) \) where \( R(n) \) is another polynomial with integer coefficients.
Rollout 91, Act: 3.805
)\).\n\n3. **Calculations**:\n - **Angle \(B\)**:\n - \
Rollout 56, Act: 3.803
3} \), and \( A_1A_2 = 2 \). The area of this
Rollout 16, Act: 3.796
^k}}\n\]\n\nCalculating each term separately:\n\n- For \( k=1 \): \( \
Rollout 96, Act: 3.793
94}{2197}\n\]\n\nFinally, summing \(\alpha\) and \(\
Rollout 31, Act: 3.760
+ xy) a_m = 0\n\]\nSince this must hold for all \( a_0
Rollout 44, Act: 3.733
\Delta U \) is negative, not positive. Statement [D] is incorrect.\n\nFinal Answer:
Rollout 44, Act: 3.722
pressure results in more work done on the gas. Therefore, statement [A] is correct.\n\n**[
Rollout 95, Act: 3.716
3}k}{490}\).\n - For this length to be rational, \(\sqrt
Rollout 83, Act: 3.712
rotations by 0 to 8 positions). However, only the identity rotation (0 positions) fixes
Rollout 40, Act: 3.698
greatest integer less than or equal to \(|p| + |q|\).\n\nUsing complex numbers, the
Rollout 59, Act: 3.697
number of heads \(k\) are calculated as follows:\n\n- For \(k = 0\) (all
Rollout 95, Act: 3.686
sqrt{3}k\) must be rational. However, \(\sqrt{3}\) is irrational
Rollout 21, Act: 3.658
. If \( f \) is the floor function, then \( x - \lfloor x \rf
Top 16 Negative Activations
Rollout 89, Act: -6.478
we give in terms of 66.666..., but if being in primary school, the
Rollout 73, Act: -6.499
98. Therefore, answer is 398.\n\nBut let's confirm with another small case.
Rollout 8, Act: -6.510
0 different cards, hence the number of ordered sequences is 10P5. If they are ind
Rollout 89, Act: -6.512
0 divided by 3, so about 33.333... so 2/5
Rollout 89, Act: -6.516
3, so units=33.33. Alternatively, girls= 2*33
Rollout 8, Act: -6.561
5 cards in order would be 10P5, as earlier.\n\nBut the problem here is that
Rollout 89, Act: -6.600
100 /0.6=166,666...\n\nBut girls are 0
Rollout 89, Act: -6.605
5)/3 ��166.666. But how can that be?\n\nWait,
Rollout 89, Act: -6.698
, girls= 2*33.33=66.666.\n\nHmm,
Rollout 8, Act: -6.703
the number of ordered sequences is 10P5. If they are indistinct, then the
Rollout 89, Act: -6.752
3. Alternatively, girls= 2*33.33=66.666
Rollout 89, Act: -6.907
0 divided by 3, so units=33.33. Alternatively, girls= 2
Rollout 76, Act: -6.911
From four subgrids, the number of pairs is C(4,2)=6. But not
Rollout 89, Act: -6.924
part is 100/3≈33.333... Therefore, 2 parts
Rollout 89, Act: -7.027
of girls would be approximately 66.67. But practically, it's around 67
Rollout 95, Act: -7.073
25/3 �� 8.333...\n\nThen x = 25/3

DOWN_PROJ

Top 16 Positive Activations
Rollout 22, Act: 8.616
2016 kg/mol. Then, divide by Avogadro's number: 0.0
Rollout 80, Act: 7.647
11/(86400) ��1.157e6 days ��
Rollout 71, Act: 7.311
, the sum of the products of the roots taken two at a time: ab + bc + ac =
Rollout 22, Act: 7.188
.016e-3 kg/mol) / 6.0221407
Rollout 54, Act: 7.015
is a stronger acid), so it's the least nucleophilic among the charged species. Methanol (
Rollout 54, Act: 6.995
aprotic solvents, alkoxide is more nucleophilic, but in polar protic, due
Rollout 75, Act: 6.984
quick check, Ramanujan proved that the only numbers requiring six positive squares are 1,
Rollout 22, Act: 6.971
02016 kg/mol. Divided by Avogadro's number (6.02
Rollout 40, Act: 6.935
+1 ) ] * (1 + i)\n\nWait, no. Wait, the term was 1
Rollout 22, Act: 6.883
3 kg/mol) / 6.02214076e23 mol−
Rollout 22, Act: 6.807
0.002016 / 6.02214076
Rollout 54, Act: 6.790
actually, steric hindrance might make it less nucleophilic because the lone pairs are harder to access
Rollout 37, Act: 6.783
3 - 3^3*5^2 = 9*125 - 27
Rollout 54, Act: 6.757
be:\n\n5. Ethanethiolate (most nucleophilic)\n\n2. Hydroxide\n\n1
Rollout 57, Act: 6.754
U around x with closure contained in A. But hold on, in the density topology, closures might not
Rollout 22, Act: 6.730
.022e23 is 0.002016 / 6.
Top 16 Negative Activations
Rollout 87, Act: -6.075
)) �� (ρ / r) dV, where �� is the charge density, r is
Rollout 62, Act: -6.079
by 121.\n\n121*143= 17303.
Rollout 56, Act: -6.097
.112 +... converges.\n\nIf we assume each difference is multiplied by 1/2
Rollout 38, Act: -6.133
)/31= (42k +888k )/31=93
Rollout 25, Act: -6.179
, then technically, the gravitational parameter would be μ = G*(M_earth + M_asteroid
Rollout 56, Act: -6.209
/12 ) = 7√3 /144\n\nSo total:\n\n= | 0
Rollout 8, Act: -6.248
4/3024=744÷24=31; 302
Rollout 73, Act: -6.293
remainder is 398. However, I should check the answer more carefully. Wait, but before
Rollout 76, Act: -6.324
29.\n\nBut hold on, before finalizing, check the arithmetic once more.\n\nTotal colorings with
Rollout 48, Act: -6.429
\n\n5 +6= B\n\n4 + B = F (hex). So for the sum, it
Rollout 49, Act: -6.504
Alternatively, use the fact that X has a countable extent (every closed discrete subspace is countable
Rollout 59, Act: -6.516
92= 515.\n\nBut let's check, some other way. Maybe think of generating
Rollout 62, Act: -6.602
7= 7*297=2079. 2081-2
Rollout 53, Act: -6.654
that in a triangle, the centroid divides each median into a ratio of 2:1, with the
Rollout 87, Act: -6.690
+ z2)] dx\n\nTo compute this integral, I can use substitution. Let’s set u =
Rollout 53, Act: -7.101
ians intersect at centroid P, which divides each median into ratio 2:1.\n\nGiven that PD =

Layer 56

GATE_PROJ

Top 16 Positive Activations
Rollout 51, Act: 9.910
(0 - 0)/n = 0\n\nTherefore, cos7x and cos9x integrated
Rollout 31, Act: 9.664
, and y=(1+√5)/2.\n\n**Final Answer**\nThe possible values of \( y
Rollout 17, Act: 9.600
3 [ 2/3 ] - 2\n\n= 2 - 2 = 0\n\n
Rollout 76, Act: 9.521
Hence, the answer is 929.\n\n**Final Answer**\n\boxed{929
Rollout 59, Act: 9.505
*7^2. No common factors. Therefore, m =123, n=39
Rollout 16, Act: 9.361
104 - 112)/7 = (-8)/7 ��-1.1
Rollout 61, Act: 9.360
10 �� 1 mod 9\n\nSo the multiplicative order of 2 mod
Rollout 13, Act: 9.352
to annual is $187.12.\n\n**Final Answer**\n\boxed{187
Rollout 59, Act: 9.347
=123, n=392.\n\nSo 123 + 392
Rollout 61, Act: 9.336
correct, leading to the answer 913.\n\n**Final Answer**\n\boxed{913
Rollout 90, Act: 9.297
the first k digits, M is the middle digit (0 ≤ M ≤ 9), and L is
Rollout 78, Act: 9.293
49. The answer is 49.\n\n**Final Answer**\n\boxed{49}\n\n
Rollout 62, Act: 9.277
121 is 11×11, so two prime factors. Okay, so now we
Rollout 32, Act: 9.266
of the entries is 4018. So that's probably the answer.\n\nWait, let's
Rollout 39, Act: 9.264
me check.\n\n10^{322} is 1 followed by 322 zeros.
Rollout 69, Act: 9.261
12 +192=204.\n\nSo, yes, conclusion is a=12
Top 16 Negative Activations
Rollout 33, Act: -4.882
's k) inserting the numbers appropriately. But perhaps not.\n\nAlternatively, maybe I can use the principle of
Rollout 95, Act: -4.893
??? Wait.\n\nWait, hold on. Let me recast this.\n\nPoint Pk is along AB.
Rollout 24, Act: -4.913
perhaps using extremal principles or inequalities?\n\nBut not sure.\n\nAlternatively, think of R as selectors for the
Rollout 64, Act: -4.951
expanding the coefficient of a1:\n\nBut perhaps not helpful.\n\nAlternatively, divide both sides by 2^{
Rollout 69, Act: -4.983
tan(3 theta) expression, but that might not help here.\n\nAlternatively, since the problem involves an
Rollout 75, Act: -5.003
may complicate things. Wait, no, perhaps not.\n\nAlternatively, notice that 8N +3
Rollout 57, Act: -5.016
with closure in the density topology. But this seems difficult.\n\nAlternatively, take two non-empty closed sets in
Rollout 93, Act: -5.073
, which complicates matter.\n\nAlternatively, it might not be useful directly because the relations depend on parity of
Rollout 49, Act: -5.124
plus metrizable implies second countable? Maybe not the direction here.\n\nWait, but our space X
Rollout 1, Act: -5.156
orthocentric system or something similar?\n\nLet me think.\n\nSuppose we fix y. Then the conditions
Rollout 66, Act: -5.159
to assign a and b parity yet. Let's proceed.\n\nSo, starting with ab = 6(a
Rollout 1, Act: -5.186
2)} / 2.\n\nBut perhaps it's hard.\n\nWait, back to the equation:\n\nFor y
Rollout 91, Act: -5.218
Hmm. Let's think.\n\nBut perhaps it's better to assign coordinates. Let me place the triangle in
Rollout 49, Act: -5.245
something?\n\nBut I’m not sure. Let me think. If every subset of X is Fσ,
Rollout 14, Act: -5.420
I to F and K.\n\nNo, that seems not immediately helpful. \n\nWait, have you consider
Rollout 64, Act: -5.445
, expanding the coefficient of a1:\n\nBut perhaps not helpful.\n\nAlternatively, divide both sides by 2

UP_PROJ

Top 16 Positive Activations
Rollout 28, Act: 8.933
(3)^2*(3)^2=0 ?\n\nWait no, with a=3, b=
Rollout 67, Act: 8.687
3936 +1020.\n\nCalculate:\n\n-72*64= -
Rollout 96, Act: 8.454
20 -4106= something wrong.\n\nWait, no:\n\nWait, 410
Rollout 96, Act: 8.428
9 +1134)/169.\n\nWait, in previous calculations, OB_x was (-
Rollout 0, Act: 8.307
2,3,5,7 are even?\n\nWait floor operations:\n\nFor prime 2:
Rollout 59, Act: 8.296
16*24.5? Wait, 16 *24.5 is
Rollout 36, Act: 8.278
688, Số này dài quá. Wait, 18,864 squared: Let
Rollout 53, Act: 8.239
1971 divided by 146:\n\n1971 ��146:\n\n
Rollout 70, Act: 8.234
51 is a 7-10 double.\n\n51 in base-7 is 10
Rollout 96, Act: 8.182
7+6+9+3=34?\n\nWait, wait:\n\n5+8=13
Rollout 36, Act: 8.120
5,688, Số này dài quá. Wait, 18,864 squared
Rollout 2, Act: 8.103
999*10 +9, wait, perhaps 9999=9*1
Rollout 71, Act: 8.076
92+492 +102. Let me compute that:\n\n392=
Rollout 96, Act: 8.011
740 / 169.\n\nWait, let me compute that. Divide numerator and denominator by
Rollout 59, Act: 8.005
4.5 is 392? No, 16 *24 is 38
Rollout 22, Act: 7.991
by 0.006333.\n\nCompute 18,168.7
Top 16 Negative Activations
Rollout 69, Act: -5.554
10^\circ}{3^i}\right) - \sec\left(\frac{30
Rollout 28, Act: -5.569
(-k a)/(b(2h -a)) - h b/(a(2k -b
Rollout 29, Act: -5.570
the handle is plugged, preventing flow, then this can't happen. Therefore, how does this affect the
Rollout 12, Act: -5.571
K}{dk} = \frac{E(k) - (1 - k^2)K(k
Rollout 69, Act: -5.573
\left( \sec(10^\circ) - 1 \right).\n\]\n\nHowever,
Rollout 24, Act: -5.637
pair (R, C), the set R has to be a minimal covering set for the rows with respect
Rollout 24, Act: -5.750
R. In a minimal pair, these partial orders must have certain properties that fix their sizes.\n\nAlternatively,
Rollout 14, Act: -5.843
sqrt{J}}{2\sqrt{A}} - \frac{K}{A^2}
Rollout 28, Act: -5.918
2 k a2 / (2h -a) - 2 h b2 / (2k
Rollout 69, Act: -5.954
\left( \sec(10^\circ) - \lim_{i \to \infty
Rollout 87, Act: -5.960
3 \ln(1 + \sqrt{2}) \right] }\n\nSince I recall this from internet
Rollout 69, Act: -6.008
( \frac{2}{\sqrt{3}} - 1 \right) = \frac{
Rollout 69, Act: -6.153
}{4} \left( \sec(\theta) - \sec(3\theta) \right
Rollout 91, Act: -6.221
| \frac{t + 1}{t} - \frac{1 - t}{t}
Rollout 30, Act: -6.234
frac{b_n}{a_{n+1}} - \frac{1}{a_{n+
Rollout 66, Act: -6.539
:\n \[\n \frac{ab}{c - a} = 6p\n \

DOWN_PROJ

Top 16 Positive Activations
Rollout 17, Act: 10.382
let's think. In electromagnetism, the magnetic field of a dipole can be written as:\n\n
Rollout 40, Act: 8.877
. Since 36π is just 18 full circles, equivalent to 0. So e
Rollout 17, Act: 8.773
First, I need to recall the formula for the magnetic field of a dipole. The magnetic field **
Rollout 6, Act: 8.755
with a sparse oracle. (This is the Meyer, I think, theorem.) So perhaps this is the
Rollout 68, Act: 8.753
ence, the desired answer is501.\n\n**Final Answer**\n\boxed{501}\n\n
Rollout 4, Act: 8.753
percentage error is approximately 0.88%.\n\n**Final Answer**\nThe percentage error in the calculation of
Rollout 17, Act: 8.511
for the magnetic field of a dipole. The magnetic field **B**(r) produced by a dip
Rollout 17, Act: 8.442
\times 10^{-2}} T.\n\n**Final Answer**\n\boxed{2.54
Rollout 8, Act: 8.261
+126= 157.\n\n**Final Answer**\n\boxed{157}\n\n
Rollout 61, Act: 8.206
, leading to the answer 913.\n\n**Final Answer**\n\boxed{913}\n\n
Rollout 47, Act: 8.110
the answer N = 9. Looks good.\n\n**Final Answer**\n\boxed{9}\n\n<|im_start|>answer
Rollout 42, Act: 8.110
hydrogen from the cyclopropane (which is highly strained) could transfer to the carbocation. But
Rollout 27, Act: 8.060
8.0×10^6 K.\n\n**Final Answer**\nThe rms speed of the Be nuclei
Rollout 76, Act: 7.969
ence, the answer is 929.\n\n**Final Answer**\n\boxed{929}\n\n
Rollout 59, Act: 7.965
Hence the answer is 515.\n\n**Final Answer**\n\boxed{515}\n\n
Rollout 0, Act: 7.932
2^{8-1}=128.\n\n**Final Answer**\n\boxed{128}\n\n
Top 16 Negative Activations
Rollout 97, Act: -8.187
stoichiometric coefficient. Hmm, no. Maybe step-by-step.\n\nLet me model the partial pressures as
Rollout 16, Act: -8.233
73...\n\nNow, adding up all these terms term by term:\n\nStart with 12 (from
Rollout 27, Act: -8.237
is confusing. Wait, let's try to move step by step. The problem first asks to calculate V
Rollout 39, Act: -8.369
more precise way is:\n\nLet's perform the division step by step. Consider that 999...
Rollout 67, Act: -8.394
=348 *a*d.\n\nWait, step by step:\n\nMultiply equation 3 by a*d
Rollout 38, Act: -8.427
this is getting a bit confusing. Let's approach step by step.\n\nFirst, define coordinate system: Let
Rollout 16, Act: -8.682
/2)^k). Let's compute this again term by term.\n\nk=1:\n\nC(6
Rollout 53, Act: -8.800
where calculations can be performed. Let's try this step by step.\n\nLet me set coordinates:\n\nLet’s
Rollout 28, Act: -8.929
2*(3n)^2=0\n\n compute term-wise:\n\n4*9n2h2=
Rollout 17, Act: -9.129
29e-11)^3 )\n\nCompute step by step:\n\nNumerator:2*4*
Rollout 83, Act: -9.212
considering symmetries.\n\nBut let me take it step by step.\n\nFirst, how many assignments in total
Rollout 45, Act: -10.250
(-3b) +0 +4.\n\nCompute term by term:\n\n25 +6a +2
Rollout 11, Act: -10.263
1/(a2 z )* a\n\nSimplify term by term:\n\n- a\n- a z\n
Rollout 51, Act: -10.426
+ 1/7 ] = let's compute term by term.\n\nFirst term inside first brackets:
Rollout 19, Act: -10.529
is 0.\n\nWait, let me do this step by step. Let's compute \( (k -
Rollout 50, Act: -10.587
0}101] \]\n\nSimplify term by term:\n\n- \((\log_{1

Layer 57

GATE_PROJ

Top 16 Positive Activations
Rollout 12, Act: 9.226
, and when x =1, t= ν.\n\nBut then x = t/ν, dx =
Rollout 12, Act: 9.057
- t2) ) / t ] dθ\n\nTherefore, dt / [ sqrt( (t2
Rollout 12, Act: 8.931
t2) ) ] = dθ / t\n\nSo, rearranged, the integrand dt /
Rollout 73, Act: 8.858
so for i from 0 to j-1. So for each j, 2^j appears
Rollout 56, Act: 8.800
in the triangle, we need to figure out coordinates.\n\nWait, actually, starting with a right-angled
Rollout 69, Act: 8.784
3 -3√3 ) /12√3 ) = (6 -3√3 )/(
Rollout 12, Act: 8.782
nu /k^2 ) / ( nu2 )\n\n= [ E(k) - nu2 K(k
Rollout 96, Act: 8.773
and 2148=-12k.\n\nFrom first equation, k=-895/
Rollout 51, Act: 8.719
dx = [-cos(nx)/n]0^pi= (-cos(n pi)/n + cos0
Rollout 73, Act: 8.708
: sum should be 1*2 =2.\n\nPlug in n=1: (1 -1
Rollout 30, Act: 8.680
_{n+1} -1 ) b_n.\n\nSo,\n\nb_n = a_{n+3
Rollout 87, Act: 8.675
1 + r2), t dr = r dt.\n\nWait, previous substitution. Since t = sqrt(
Rollout 17, Act: 8.608
/T, and π a03 is m3.\n\nCalculating units:\n\n( T·m/A *
Rollout 51, Act: 8.583
0 + sin8θ*(-4/63) = -4/63 sin8θ.\n\n
Rollout 17, Act: 8.577
μ_p ) / (3 π a03 )\n\nSo in numbers:\n\n2*(4πe-
Rollout 8, Act: 8.570
,k) * P(5,5 -k)]\n\nWait, for k=1:\n\nNumber for each
Top 16 Negative Activations
Rollout 21, Act: -4.839
Proposed by Carl Schildkraut*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so part (
Rollout 7, Act: -4.843
of non-promoted $47$ .\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I'm
Rollout 94, Act: -4.845
,d\n(D) a,b\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to determine which
Rollout 54, Act: -4.846
3, B = 2\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's tackle this question step
Rollout 43, Act: -4.848
imes\left|0\right\rangle\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try
Rollout 16, Act: -4.850
.\n\n*Proposed by Yannick Yao*\n<|im_start|>assistant\n<|im_start|>think\nOkay, so Emilia
Rollout 97, Act: -4.861
of $\mathrm{Y}$?\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's see. I have
Rollout 27, Act: -4.898
of velocity along the line of sight).\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I've got this
Rollout 3, Act: -4.944
radon (Rn) in m/s\n<|im_start|>assistant\n<|im_start|>think\nOkay, I need to
Rollout 33, Act: -4.965
permutation is greater than $k$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to find
Rollout 1, Act: -4.966
$ \mathcal{H}$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, so I need to show
Rollout 32, Act: -4.969
M}^{2009}.$\n<|im_start|>assistant\n<|im_start|>think\nAlright, so I need
Rollout 24, Act: -4.991
contain the same number of rows.\n<|im_start|>assistant\n<|im_start|>think\nAlright, let me start by trying to
Rollout 17, Act: -5.017
}|$ (in SI units).\n<|im_start|>assistant\n<|im_start|>think\nAlright, let me try to tackle this
Rollout 26, Act: -5.036
~mol}^{-1}$ ]\n<|im_start|>assistant\n<|im_start|>think\nOkay, let's try to figure this
Rollout 66, Act: -5.070
$ . Find $p$ .\n<|im_start|>assistant\n<|im_start|>think\nOkay, let me try to work through

UP_PROJ

Top 16 Positive Activations
Rollout 9, Act: 8.191
to make this clearer.\n\nLet's place the square on a coordinate system. Let's say point A is
Rollout 91, Act: 8.014
better to assign coordinates. Let me place the triangle in coordinate system. Let me set point B at (-
Rollout 95, Act: 7.818
it's a right triangle? Let me check using the Pythagorean theorem. If ABC is a right
Rollout 38, Act: 7.803
between southwest and southeast corners.\n\nWait, let's build a coordinate system. Let’s consider the northeast corner
Rollout 9, Act: 7.759
on the left side of EM. Wait, in our coordinate system, EM is at x=10
Rollout 9, Act: 7.665
make this clearer.\n\nLet's place the square on a coordinate system. Let's say point A is at
Rollout 2, Act: 7.643
? mod 1000\n\nSolve this system using Chinese Remainder Theorem.\n\nWe need
Rollout 11, Act: 7.610
must have non-real coefficients, making it hard to apply that.\n\nAlternatively, think of log variables? If
Rollout 38, Act: 7.532
+336=0\n\nNow, we have quadratic in r:\n\n336 r2 -
Rollout 53, Act: 7.525
.\n\nAlternatively, it's possible to place this triangle into coordinates where calculations can be performed. Let's try
Rollout 59, Act: 7.437
515.\n\nBut let's check, some other way. Maybe think of generating functions?\n\nThe
Rollout 58, Act: 7.405
171}.\n\nBut let me see if that makes sense. Let me do a smaller case.
Rollout 81, Act: 7.402
1 = 0\n\nLet's attempt to solve this quadratic equation for h. Quadratic formula: h
Rollout 81, Act: 7.369
3, DC=17. Then in such a case, sometimes, in certain triangles with given ratios
Rollout 29, Act: 7.367
a higher pressure at the bottom.\n\nWait, this seems contradictory. Let's make it clearer. If the
Rollout 37, Act: 7.366
monde matrices look like [[1, x, x^2, ..., x^{n-1}],
Top 16 Negative Activations
Rollout 75, Act: -4.444
42 + 32. That works. Similarly, 32=9: 9
Rollout 93, Act: -4.459
. Hence sum=1007 as well. Wait, j can go up to201
Rollout 71, Act: -4.529
So d≈25 or d≈26. But d can be up to 25 since
Rollout 71, Act: -4.581
by primes up to sqrt(2011). sqrt(2011) is approx
Rollout 38, Act: -4.627
6) is less than 1055. Compute 10542=(10
Rollout 90, Act: -4.632
12108 is 5 digits. The middle digit is the third one, which is
Rollout 3, Act: -4.684
root.\n\nsqrt(28,436). Let's see, 168^2
Rollout 3, Act: -4.712
^2 = 28,224. 169^2 = 28
Rollout 22, Act: -4.712
2,890,000. Therefore, sqrt(2,869,
Rollout 51, Act: -4.748
[sinn pi - sinn0]/n. But since n is integer, sinn pi is
Rollout 19, Act: -4.896
, its divisors are 1 and83. Hence, only divisors ≥6 are83
Rollout 19, Act: -4.900
of 83 greater than or equal to6. Since 83 is prime, its divisors
Rollout 90, Act: -4.916
would result in B being 1090. Is 1090 divisible by 2
Rollout 71, Act: -4.935
4-300=7744. 7744=882.
Rollout 72, Act: -4.998
), the only root of unity is 1 itself. So we need to add three 1s to
Rollout 71, Act: -5.107
1 -1875=136. 136 is not a perfect square.\n\n

DOWN_PROJ

Top 16 Positive Activations
Rollout 25, Act: 10.921
�� 6.8e-5 m/s2. Even such a massive asteroid (1e
Rollout 49, Act: 10.312
arily Fσ" or something?\n\nBut I’m not sure. Let me think. If every subset of
Rollout 49, Act: 10.311
ixson decomposition is countable. But I am not sure.\n\nAlternatively, think about the fact that if
Rollout 89, Act: 10.262
. If they use fractions in boys. Maybe they just require an exact answer even if not a whole number
Rollout 28, Act: 10.215
known as the nine-point ellipse. But I'm not exactly sure. It requires more knowledge.\n\nIn any
Rollout 16, Act: 10.177
tree. Each sapling independently succeeds with probability 1/2 each day. So the process continues until
Rollout 93, Act: 10.085
an integral over some function, but again I'm not sure.\n\nWait, let's try a small case
Rollout 28, Act: 9.845
feet of altitudes and centroids... but I'm not sure.\n\nAlternatively, see that the locus is known
Rollout 28, Act: 9.841
conics are hyperbolas because the discriminant B2 -4AC>0 for all non-de
Rollout 49, Act: 9.834
Or maybe has countable extent? But I'm not sure.\n\nAlternatively, in a hereditarily Lind
Rollout 49, Act: 9.664
and Hausdorff implies regular? I'm not sure.\n\nWait, actually, a hereditarily
Rollout 49, Act: 9.658
right-separated spaces are countable). But I'm not sure.\n\nHere's a theorem: every Lindel
Rollout 27, Act: 9.580
rms fluctuation in energy would be sqrt(⟨E2⟩ - ⟨E⟩2). Let's
Rollout 5, Act: 9.532
areas, and then find the greatest integer not exceeding x2/100. \n\nFirst, let
Rollout 22, Act: 9.477
divided by kg/mol gives (kg·m2/s2)/kg) = m2/s2.
Rollout 71, Act: 9.392
add up to -c, and their product is c2 - 2011. Therefore,
Top 16 Negative Activations
Rollout 49, Act: -8.443
, hence Borel. But Bernstein sets are not Borel. Therefore, such a Bernstein set
Rollout 49, Act: -8.451
ff, uncountable), there are subsets that are not Fσ. For example, the set of
Rollout 49, Act: -8.463
uncountable closed set. Such a set is not Borel, hence not Fσ. But
Rollout 65, Act: -8.490
1995*q. Since p and q are coprime, then p divides199
Rollout 49, Act: -8.491
every subset is Fσ, then every subset is a countable union of closed sets. Therefore, the
Rollout 16, Act: -8.569
be in the form m/n where m and n are coprime, and then compute 10
Rollout 49, Act: -8.572
, in complete metric spaces, the irrationals are a Gδ set, but complement of Gδ
Rollout 49, Act: -8.701
every subset is Fσ, then every subset is a countable union of closed sets, each of which
Rollout 49, Act: -8.786
is Fσ, then every subset is a countable union of closed sets. Therefore, the Bore
Rollout 36, Act: -8.826
92}.\n\nBut original problem seems to originate from a competition likely answer is integer, which 1
Rollout 54, Act: -8.832
basic, hydroxide is smaller and may engage in hydrogen bonding with solvent more. But maybe the lower
Rollout 49, Act: -8.979
. For example, the set of irrationals is not Fσ. Wait, actually, in complete
Rollout 57, Act: -8.984
be separated. As a classical example, consider two disjoint Lebesgue measure zero sets. There exist two
Rollout 49, Act: -9.320
, consider a Bernstein set. A Bernstein set is uncountable, meets every perfect set, so it
Rollout 49, Act: -9.407
Cantor set are not Fσ, like the co-Fσ sets. For example, a Gδ
Rollout 57, Act: -10.533
As a classical example, consider two disjoint Lebesgue measure zero sets. There exist two dense measure zero

Layer 58

GATE_PROJ

Top 16 Positive Activations
Rollout 13, Act: 3.569
[\n1.12^2 = 1.2544 \quad \text{and
Rollout 42, Act: 3.065
. The hydroxyl group is protonated, forming a good leaving group (water). Upon departure of
Rollout 13, Act: 2.835
\n1.12^2 = 1.12 * 1.12. Let
Rollout 42, Act: 2.752
typically undergo E1 mechanisms because they can form more stable carbocations.\n\nSo the initial step would be proton
Rollout 6, Act: 2.711
Meyer's theorem equivalence between \( \textbf{P}/\text{poly} \) and sparse
Rollout 13, Act: 2.708
�� 6000 * 1.57351936 ��
Rollout 42, Act: 2.706
aqueous H3PO4, which is a good acid, it's more likely an E1 mechanism
Rollout 42, Act: 2.696
group is protonated, forming a good leaving group (water). Upon departure of water, a **ter
Rollout 42, Act: 2.691
into a good leaving group (water). Then, the elimination would occur, removing a proton (H+)
Rollout 42, Act: 2.674
O+), it leaves as water, generating a carbocation. Wait, but if the mechanism is E
Rollout 42, Act: 2.647
utane-derived) ring. \n\nUltimately, elimination of a beta-hydrogen results in the formation of
Rollout 23, Act: 2.607
of the 115 Bq gives3.17%×115≈3.
Rollout 42, Act: 2.607
ane-derived) ring. \n\nUltimately, elimination of a beta-hydrogen results in the formation of a
Rollout 54, Act: 2.567
.\n\nRegarding part B:\n\nWe have to choose statement 1 or 2.\n\nStatements:\n\n1. In
Rollout 42, Act: 2.454
the cyclopropane can act as a conjugated diradical and combine with the carbocation.
Rollout 20, Act: 2.447
modulo43. Since2 has multiplicative order 14 mod43 (as φ(4
Top 16 Negative Activations
Rollout 71, Act: -8.369
roots of the quadratic x2 + cx + (c2 - 2011). Because if
Rollout 64, Act: -8.378
+36 ? Not sure.\n\nAlternatively, the answer is 2^{2}*3^{2}(
Rollout 11, Act: -8.388
to the sum of the variables plus the sum of their products two at a time. That form makes me
Rollout 17, Act: -8.410
/s2) ) / T? Wait, maybe it's simpler to note that units check out:\n\n**
Rollout 28, Act: -8.462
2/a2). So the coefficients are:\n\nFor x2: a2/b2\n\nFor xy:
Rollout 69, Act: -8.479
1}\n\nAlternatively, let's suppose we have\n\nLet’s define some recurrence here. Suppose we set x
Rollout 59, Act: -8.531
squares of each probability:\n\n(3/28)2 + (10/28)2
Rollout 75, Act: -8.568
all distinct and odd. We need to show that a2 + b2 + c2 can be written
Rollout 16, Act: -8.568
3.\n\nWait, but this seems large. The answer is 789,953?
Rollout 71, Act: -8.585
add up to -c, and their product is c2 - 2011. Therefore,
Rollout 51, Act: -8.633
's compute Integral A and Integral B.\n\nSo, let's compute Integral A: ��02π
Rollout 16, Act: -8.771
2, ..., T6}. So the expected value we need is E[T].\n\nSo, we can use
Rollout 69, Act: -9.125
about manipulating the term specifically. Let's write:\n\nLet’s look at the term term_i = sin2
Rollout 69, Act: -9.214
using the tangent function.\n\nWait, another idea. Let me note that 3 theta_i = theta_{
Rollout 95, Act: -9.248
if the right angle is at B, but 92 + (5√3)2 =
Rollout 69, Act: -9.285
, let me think about manipulating the term specifically. Let's write:\n\nLet’s look at the term term

UP_PROJ

Top 16 Positive Activations
Rollout 64, Act: 5.039
*(2^{2009} -1 ) +8 -1=8*(2^{2
Rollout 7, Act: 4.710
56N -56P=56N +15P=66N\n\nSo
Rollout 18, Act: 4.646
-5 -25k)/3= (0 -9k)/3= -3k.\n\n
Rollout 62, Act: 4.559
-m +11 ��m2 -m +1 ��0 mod5.\n\nSo m
Rollout 11, Act: 4.515
5/4) e^{i2θ} - (15/4) e^{-i2
Rollout 81, Act: 4.473
*400 +2601*84=1,040,40
Rollout 75, Act: 4.428
4 + 4 + 1 + 9 + 1 = 35\), which uses
Rollout 28, Act: 4.356
hk + b2 h2 +a2 k2 - a b2 h - a2b k
Rollout 67, Act: 4.349
2 d q ) ] n + [ a p - a q - p d + d q ]\n\n
Rollout 81, Act: 4.321
/3 - h/17 )/( 1 + (h/17)(-h/
Rollout 2, Act: 4.312
999=999*10 +9, wait, perhaps 9999
Rollout 90, Act: 4.297
2018*(300 +80 +8)=2018*30
Rollout 64, Act: 4.245
1 ] / (2^{2009} -1 )=8 +7/(2^{
Rollout 62, Act: 4.241
-1 +11=121k2 +11k +11=11
Rollout 28, Act: 4.178
) k2 -4b h -4a k +ab=0\n\nAssuming a and b
Rollout 28, Act: 4.170
(-k a)/(b(2h -a)) - h b/(a(2k -b
Top 16 Negative Activations
Rollout 99, Act: -7.610
and "what's the decimal value of x3".\n\nWait, but starting from x0, x1
Rollout 72, Act: -7.613
but order n, which usually means a positive integer.\n\nTherefore, the answer is all even positive integers.
Rollout 58, Act: -7.625
37,171. That's correct.\n\nBut we need to make sure that the result is
Rollout 43, Act: -7.646
to |1⟩ when control is |0⟩.\n\nTherefore, the output is ( alpha |0⟩
Rollout 15, Act: -7.651
all lamps will be off after some time.\n\nWait, but I need to be careful here. The problem
Rollout 51, Act: -7.662
However, sin8θ still is a real number.\n\nBut maybe we can write absolute value? Wait,
Rollout 19, Act: -7.697
Therefore, only.Therefore, n=78.\n\nBut in the problem statement, in the original modulus
Rollout 88, Act: -7.723
channels into a vector, referring to weights. Hmm.\n\nWait, let me check again: "What is
Rollout 99, Act: -7.734
1. What's the decimal value of x3?"\n\nWait, each iteration uses the last three points.
Rollout 59, Act: -7.743
get squared as (3/28)^2?\n\nWait, but wait, so originally each of their
Rollout 44, Act: -7.744
every step, hence maximum work is done. Wait, no: wait, in compression, the work done
Rollout 76, Act: -7.745
, the squares in their union must all be red.\n\nYes, because for S1 to be all red
Rollout 55, Act: -7.777
/2], with the middle entry as 1.\n\nLet me write that out:\n\n\[\n\math
Rollout 43, Act: -7.869
. So yes, it's control followed by target.\n\nThus, answer C is correct. Then the mistake
Rollout 43, Act: -8.049
control (first qubit) is |0⟩.\n\nSo first qubit is control, second is target
Rollout 73, Act: -8.129
0 term which is 0, so same result.\n\nBut perhaps I should confirm.\n\nAlternatively, I recall

DOWN_PROJ

Top 16 Positive Activations
Rollout 92, Act: 13.999
/61.5≈0.016260\n\nSubtract: 0.
Rollout 25, Act: 13.421
12.714e6≈6.85e14\n\nThis is approximation for
Rollout 56, Act: 13.280
1/1008 ��1.827\n\nThen * ��3≈1
Rollout 25, Act: 13.279
1.2714e7≈6.849e14 m^3/s
Rollout 92, Act: 13.062
.4054) - (1.6260) = 3.77
Rollout 53, Act: 13.043
=414/73 ��5.67.\n\nSimilarly, sum over x_i y
Rollout 92, Act: 12.904
.615) is approximately -1.62601626. So adding
Rollout 49, Act: 12.775
In general, every subset of a compact metric space is σ-compact? No. The space {0
Rollout 3, Act: 12.328
98) ) �� sqrt( ~225,753 ) �� 4
Rollout 99, Act: 11.950
6315) �� 1.1573\n\nThus x3 ��
Rollout 99, Act: 11.932
486715)≈1.15733466671
Rollout 49, Act: 11.805
Alternatively, consider that in X, every subspace is σ-compact, so in particular, X itself
Rollout 92, Act: 11.208
54054 -0.01626 = 0.0378
Rollout 99, Act: 11.190
4563) �� 1.1575\n\nTherefore, x3 = x
Rollout 99, Act: 11.169
(-3.4563)=1.1573.\n\nThus, x3=x2
Rollout 3, Act: 11.127
/0.697433≈26,034.21.\n\nsqrt
Top 16 Negative Activations
Rollout 62, Act: -7.941
is called a sphenic number if it's square-free with three distinct primes, but here it can
Rollout 25, Act: -7.980
negative. Hence, μ_final is negative, which is impossible. That suggests that given these parameters (original
Rollout 95, Act: -8.016
conclusion after detailed calculation. Therefore, the answer is \boxed{0}。\n\n**Final Answer**\n\
Rollout 26, Act: -8.020
083 * 298 �� 0.0083 * 30
Rollout 55, Act: -8.064
�� 12 = 168.166..., so 168*
Rollout 3, Act: -8.088
but as per my knowledge cutoff is 2023, so if need to answer, in the
Rollout 66, Act: -8.198
. Therefore, the altitude is (leg1 * leg2)/hypotenuse.\n\nGiven that ABC0
Rollout 33, Act: -8.205
6*6=36\n\nk=4:24*2=48\n\nk=
Rollout 36, Act: -8.306
0\n\nThen compute 8642 = 746,496\n\nAnd cross
Rollout 82, Act: -8.327
9≈1.6279*0.623≈1.015.\n\n
Rollout 36, Act: -8.327
+ 1,728s2 + 746,496\n\nSimilarly,
Rollout 86, Act: -8.390
/8. Numerical value: pi^3≈31.006, so 3
Rollout 53, Act: -8.402
)=729/292= approx 2.496 approximately.\n\nTherefore, given
Rollout 96, Act: -8.733
down:5000*895=4,475,000\n\n
Rollout 71, Act: -8.804
, c2 >2011 => c>=45 (since 442=1
Rollout 36, Act: -9.572
37728s2 + 355690496)/(4s

Layer 59

GATE_PROJ

Top 16 Positive Activations
Rollout 20, Act: 10.294
+7 +23 +43 +47. Let me add that:\n\n2 + 3
Rollout 61, Act: 9.642
5 +19+13 +7 +1= indeed let's compute:\n\n37 +
Rollout 88, Act: 9.417
weights would be 5*5*1*4=100. The biases are additional parameters
Rollout 7, Act: 9.338
66N +5N =71N.\n\nBut separately, we can say total promoted score
Rollout 7, Act: 9.333
59(R - x) = 71N.\n\nAdditionally, we have original relations:\n\nOriginal total
Rollout 7, Act: 9.209
) = 76P + 61R = 66N +5N =7
Rollout 32, Act: 9.150
M + M2) + 669*I = 670*(M + M2
Rollout 41, Act: 9.121
0 + 252 + 576 =927.\n\nBut according to the all
Rollout 16, Act: 9.112
0*7880 +1953. Wait, but let's verify once more.\n\n
Rollout 58, Act: 9.005
i + 1))^{i} ) ] ) because product_{j≠i} (j/(
Rollout 59, Act: 8.994
^2 +11^2 +4^2=9+100+121
Rollout 67, Act: 8.951
d = -72*2 +492 =-144 +492=
Rollout 32, Act: 8.909
008} + M^{2009}\n\nBut starting from k=1 to k=
Rollout 61, Act: 8.905
of 133: 7*19=133. Check 17:
Rollout 59, Act: 8.891
/2)(1/2)(3/7) since the biased coin's probability of tails is
Rollout 75, Act: 8.890
)^2 + (2n +1)^2 = 4(k2 +k +m2 +
Top 16 Negative Activations
Rollout 49, Act: -3.458
perfect set is σ-compact. Therefore, it's a union of countably many compact sets. Each
Rollout 49, Act: -3.472
-compact space which is Lindelöf is a countable union of compact metrizable spaces...
Rollout 49, Act: -3.525
compact, then in particular, X is a σ-compact Hausdorff space.\n\nSo X =
Rollout 49, Act: -3.551
is sigma-compact if and only if it is an Fσ set. A Bernstein set is not B
Rollout 49, Act: -3.555
compact Hausdorff space with no isolated points is uncountable, right? For example, the
Rollout 49, Act: -3.560
als are completely metrizable, hence they are a Polish space, hence they are σ-compact?
Rollout 49, Act: -3.598
. Since P is σ-compact, it is a countable union of compact subsets. Each compact subset
Rollout 49, Act: -3.628
point is a Gδ set, then maybe X is countable.\n\nWait, in a Hausdor
Rollout 49, Act: -3.654
} (X - F_{m}), which is a Gδ set. It's as we had before
Rollout 49, Act: -3.676
, in complete metric spaces, the irrationals are a Gδ set, but complement of Gδ
Rollout 49, Act: -3.685
subspaces, so X is a continuous image of a certain space.\n\nThis seems too vague.\n\nAlternatively,
Rollout 49, Act: -3.743
every open set is Fσ, then the space is hereditarily Lindelöf.\n\nWait,
Rollout 49, Act: -3.890
metric spaces, σ-compactness is equivalent to being Lindelöf and second countable? Wait
Rollout 49, Act: -4.207
all points being Gδ, the diagonal is a Gδ set in the square of X.
Rollout 49, Act: -4.351
in complete metric spaces, the irrationals are a Gδ set, but complement of Gδ (
Rollout 49, Act: -4.397
points being Gδ, the diagonal is a Gδ set in the square of X.

UP_PROJ

Top 16 Positive Activations
Rollout 43, Act: 7.888
-NOT will apply an X gate (i.e., flip the target qubit) only if the control
Rollout 54, Act: 7.706
the nucleophile. However, hydroxide is a stronger base than propionate (since propionic acid
Rollout 42, Act: 7.613
cyclopropane ring adjacent to a leaving group, elimination can lead to ring opening. Alternatively, perhaps
Rollout 54, Act: 7.472
because the solvent can stabilize charge through solvation. For example, in polar protic solvents: I
Rollout 43, Act: 7.454
The standard C-NOT flips the target qubit (applies an X gate) if the control q
Rollout 54, Act: 7.401
the better it can break through solvation and attack. Also, weaker bases might be poorer nucleophiles
Rollout 93, Act: 7.372
2*1007)). Wait, is that?\n\nWait, 504π/1
Rollout 11, Act: 7.367
, then r3 +1/r3 >2 ( as AM ≥ GM, minimum at r=1
Rollout 43, Act: 7.361
). The C-NOT will apply an X gate (i.e., flip the target qubit) only
Rollout 54, Act: 7.325
less nucleophilic because the lone pairs are harder to access, but in solution, maybe also less sol
Rollout 54, Act: 7.322
comparing different anions: bulkier ones might be less nucleophilic because of steric hindrance.
Rollout 71, Act: 7.300
around 287.2857... which is not an integer. 11?
Rollout 43, Act: 7.254
the standard C-NOT. Wait, is this a C-NOT where the control is still the first
Rollout 43, Act: 7.253
instead of |1⟩? So it's like a controlled-X gate with control state |0⟩.\n\n
Rollout 69, Act: 7.215
theta? Not sure, maybe it's part of a multiple-angle identity.\n\nAlternatively, perhaps writing the term
Rollout 53, Act: 7.208
? Maybe quadrilateral AEDC is split into two triangles: AED and EDC? Wait,
Top 16 Negative Activations
Rollout 97, Act: -2.645
2O5→...), thus, the answer derived from considering the stoiChiorelative derivation is
Rollout 6, Act: -2.649
with α_n | n ∈ N }, then this set is sparse.\n\nTo retrieve α_n, the oracle
Rollout 82, Act: -2.652
021. Total symbols≈2.021.\n\nSo worse than the all-pairs but
Rollout 24, Act: -2.655
) and (R’, C’) be minimal saddle pairs.\n\nDefine a mapping f:R→R’
Rollout 46, Act: -2.655
∂U/∂x\n\nd2y/dt2 + 2 �� (dx/dt
Rollout 17, Act: -2.758
magnitude of the integral $|\mathbf{I}|$ (in SI units).\n<|im_start|>assistant\n<|im_start|>
Rollout 82, Act: -2.765
attemps:\n\nBut expected symbols≈2.021>2.\n\nSo maybe there are certain constraints
Rollout 65, Act: -2.780
(1995/c)*(19952/c)=1995^3 /
Rollout 25, Act: -2.789
x [ v^4 - (x v^4)/r_p_initial2 + 2 v^
Rollout 77, Act: -2.800
\nThe function $f$ , defined on the set of ordered pairs of positive integers, satisfies the following
Rollout 25, Act: -2.801
:\n\nr_p_final = [ (r2 v2)/μ_final ] / [1 + sqrt(
Rollout 65, Act: -2.811
32×52×72×192. The number of divisors is (2
Rollout 97, Act: -2.813
.\n\nSince pressure is proportional to moles,\n\nTotal pressure = (n0+ ~xi}P0
Rollout 97, Act: -2.814
1-2y).\n\nBut rate= d~xi/dt= k (1-2y)\n\n
Rollout 99, Act: -2.917
875086175051)=approx2.65731
Rollout 97, Act: -3.240
(initial)=moles of N2O5}= P0 V / R T}= Assuming closed vessel.\n\n

DOWN_PROJ

Top 16 Positive Activations
Rollout 54, Act: 12.741
opposite to basicity because the solvent can stabilize charge through solvation. For example, in polar protic
Rollout 57, Act: 12.208
where all open sets are measurable and have the property of containing density neighborhoods around each point. Moreover, in
Rollout 49, Act: 11.865
subset is Fσ, the space is hereditarily Lindelöf.\n\nBecause in a space where
Rollout 3, Act: 11.846
terms of molar mass, since it's a monatomic gas, the molar mass would be approximately
Rollout 49, Act: 11.541
is a stronger condition, perhaps called "hereditarily Fσ" or something?\n\nBut I’m not
Rollout 57, Act: 11.533
density point, which is precisely the definition of being "open in the density topology".\n\nWait, so this
Rollout 67, Act: 11.462
(3)=9, a=8(3)=24, p=180/3
Rollout 80, Act: 11.454
92.646 × 10^24 g = 9.264
Rollout 49, Act: 11.257
is Fσ, then the space is hereditarily a countable union of closed sets, hence in
Rollout 35, Act: 11.146
=-1. Then n=18 and n=23. Contradiction.\n\nCase 2
Rollout 6, Act: 11.104
is similar to problem 1. If you consider countably infinite advice strings for all n, and represent
Rollout 15, Act: 11.094
.\n\nL4: neighbor L3 (off). Remains off.\n\nThus, t=4: O
Rollout 62, Act: 10.864
again.\n\nm=24:576 -24 +11=576-
Rollout 24, Act: 10.849
closure operators and closed sets. Minimal pairs may correspond to meet-irreducible elements in the lattice,
Rollout 39, Act: 10.752
789 sums to7+8+9=24. Thus, total is342
Rollout 20, Act: 10.668
* 7.\n\n46 factors into 2 * 23.\n\nSo LCM(42
Top 16 Negative Activations
Rollout 58, Act: -10.081
part i are wrong is (i/(i+1))^{i - 1}. For the other
Rollout 23, Act: -10.091
6/(5.88)=0.1178837216 s−
Rollout 69, Act: -10.142
_i) = [1 - cos(2 theta_i)] / [2 cos(3 theta_i)]\n\n
Rollout 60, Act: -10.253
6, since 82*6 = 492, so 492 /
Rollout 87, Act: -10.262
(r,θ) [ (1/r) * r2 sinθ dr dθ dφ ]\n\nBut
Rollout 99, Act: -10.361
75 - (-0.39794))/(1.1575 +0.
Rollout 17, Act: -10.433
we integrate over all angles (since dV = r2 sinθ dθ dφ dr), we
Rollout 17, Act: -10.627
integrating the magnetic field over the probability density |Ψ(r)|2. Since the electron's wavefunction is
Rollout 23, Act: -10.646
1/5.88≈0.1179 s−1\n\nThen lambda_total=
Rollout 38, Act: -10.672
1 +7/24)= sqrt(1 +49/576)/ (31
Rollout 23, Act: -10.857
the previous steps that I can’t see, but each step seems correct. So likely, answer is B
Rollout 17, Act: -10.873
by the proton and the probability density |Ψ(r)|2 of the electron in the ground state of a
Rollout 17, Act: -10.957
_p] divided by r3 multiplied by |Ψ(r)|2, then integrated over all space.\n\nBut
Rollout 69, Act: -11.338
^\infty [1 - cos(2t_i)]/(2 cos(3t_i))\n\nNote
Rollout 69, Act: -11.548
our term becomes [1 - cos(2t_i)]/(2 cos(3t_i))\n\nLet
Rollout 81, Act: -11.723
BAC)= [tan(angle BAD)+tan(angle CAD)]/[1 -tan(angle BAD)tan(angle

Layer 60

GATE_PROJ

Top 16 Positive Activations
Rollout 76, Act: 1.901
that does not have a 2-by-2 red square is $\frac {m}{n}$ ,
Rollout 42, Act: 1.877
,4,5,6-hexahydropentalene\n(D) spiro[3.
Rollout 57, Act: 1.669
a density point, which is precisely the definition of being "open in the density topology".\n\nWait, so
Rollout 5, Act: 1.600
zoid into two regions whose areas are in the ratio $2: 3$ . Let $x
Rollout 42, Act: 1.538
4,5,6-hexahydropentalene\n(D) spiro[3.4
Rollout 65, Act: 1.504
and both of which have nonzero volume. Given that $b=1995,$ for how
Rollout 74, Act: 1.484
, and K_s is the coefficient for the heeling error correction. But wait, maybe K_b and
Rollout 54, Act: 1.395
-olate, 2. Hydroxide, 3. Propionate, 4. Methanol
Rollout 49, Act: 1.277
it is first countable? Or maybe has countable extent? But I'm not sure.\n\nAlternatively,
Rollout 15, Act: 1.270
". Therefore, the lamp's own state at t doesn't matter. The next state depends only on the
Rollout 32, Act: 1.262
\end{pmatrix}.\]Find the sum of the entries of $\mathbf{M}
Rollout 49, Act: 1.240
öf space, each closed set has a countable subset dense in it (since scattered implies right-separated
Rollout 49, Act: 1.193
Alternatively, use the fact that X has a countable extent (every closed discrete subspace is countable
Rollout 49, Act: 1.136
-countable, let me assume it has a countable base.\n\nWait, no. For example, the
Rollout 77, Act: 1.105
on the set of ordered pairs of positive integers, satisfies the following properties: \[f(x, x
Rollout 65, Act: 1.102
cuts $P_{}$ into two prisms, one of which is similar to $P_{},
Top 16 Negative Activations
Rollout 69, Act: -10.435
/12√3 ) = (6 -3√3 )/(12√3 )\n\nBut
Rollout 95, Act: -10.473
is at B, but 92 + (5√3)2 = 81 +
Rollout 12, Act: -10.483
t2)\n\nSo denominator becomes sqrt( ( (t2 - ν2)/ν2 ) (1
Rollout 11, Act: -10.504
1. Then abc= �� k * ��2 k *1/(k2 ��3
Rollout 69, Act: -10.506
/(12√3 )\n\nBut 6 -3√3 )/(12√3 )=
Rollout 36, Act: -10.531
^2 + z^2] = 60√10\n2. Distance from P to
Rollout 69, Act: -10.557
3 /3 -1 )/4= (2√3 -3 )/(12 ).\n\nTherefore
Rollout 17, Act: -10.565
∫_{-1}^{1} (3u2 -1) (-du) = ��
Rollout 65, Act: -10.584
1995^3/(1995^4/a2)= a2 /19
Rollout 1, Act: -10.634
= 2d / sqrt(2) = d sqrt(2) > d, so the spheres
Rollout 69, Act: -10.643
√3 )/(4*3 )= (2√3 -3 ) / 12.
Rollout 95, Act: -10.673
√3)/2 = (21 + 5√3)/2. Hmm, seems complicated.
Rollout 14, Act: -10.762
):\n\nλ2 λ=1/12→λ3=1/12→λ=(1
Rollout 95, Act: -11.061
is 12 units, and BC is 5√3. Hmm, the sides are interesting.
Rollout 32, Act: -11.233
2) + (-ω)(1) = ��4 + (-ω). But ��3 =
Rollout 17, Act: -11.463
1}^1 du = 3 [ (u3/3) from -1 to 1

UP_PROJ

Top 16 Positive Activations
Rollout 14, Act: 9.070
2 dx) * sqrt(∫12 dx) ) = sqrt(J) * sqrt(∫1
Rollout 14, Act: 8.888
} (�� x dx )^{3/2} ? But again, ��xdx diverges
Rollout 81, Act: 8.728
/ (b2 + c2 -a2) where Δ is area, but perhaps more straightforward is
Rollout 37, Act: 8.712
, which is divisible by 2015! when evaluated at integer arguments? Wait, but let
Rollout 24, Act: 8.695
in terms of the relation defined by domination. R is a minimal set such that every row is either in
Rollout 24, Act: 8.684
R.\n\nMinimality implies that R and C are minimal such dominating sets.\n\nBut in bipartite
Rollout 24, Act: 8.663
ensuring they all have the same size. For instance, in a bipartite graph with a perfect matching
Rollout 37, Act: 8.600
a_j), which in turn is divisible by n! ?\n\nWait, if we have that det(A)
Rollout 24, Act: 8.574
same sizes. Maybe use Dilworth's theorem or similar.\n\nAlternatively, think in terms of the two minim
Rollout 14, Act: 8.567
1^2 dx)(∫f^2 dx) by Cauchy-Schwarz, but as
Rollout 28, Act: 8.543
trivial where both are zero, but that gives a line, not a conic). Let's check.\n\n
Rollout 14, Act: 8.509
= sqrt(J) * sqrt(∫1 dx) = sqrt(1) *∞ =∞.
Rollout 24, Act: 8.465
R.\n\nTherefore, maybe the size of R is the minimal number of rows needed to cover all rows under
Rollout 24, Act: 8.439
C) to be a saddle pair, R must be a dominating set for all rows in the row dominance
Rollout 37, Act: 8.414
algebraic construction inherently has 2015! as a因子。 But In the algebraic scenario
Rollout 5, Act: 8.350
In a trapezoid, the midline (the segment connecting the midpoints of the legs)
Top 16 Negative Activations
Rollout 36, Act: -2.348
2\n\nFirst, compute discriminant:\n\nΔ=30,6242 -4*1
Rollout 66, Act: -2.363
, somewhere, my reasoning must be wrong.\n\nWait, but m and n must be of opposite parity.
Rollout 25, Act: -2.364
Earth, so there must be an answer. Hence, there must be an error in reasoning.\n\nWait.
Rollout 30, Act: -2.392
1}=\left(a_{n-1}-1\right)\left(b_{n}+1
Rollout 7, Act: -2.402
. So I must have made a mistake.\n\nWait, alternative approach. Let me use another angle.\n\nGiven
Rollout 7, Act: -2.411
likely, the problem expects there to be solutions. So I must have made a mistake.\n\nWait, alternative
Rollout 95, Act: -2.474
ask for such an answer if it were zero.\n\nSo, clearly, there's an error in this reasoning
Rollout 90, Act: -2.515
14 mod2018⇒L=2018-1,014=
Rollout 23, Act: -2.532
is ~8. Something's wrong here.\n\nWait, wait. An alternative approach: maybe the term "
Rollout 39, Act: -2.620
k=1}^{321} 10^k = 10 \frac{
Rollout 89, Act: -2.687
. The problem wouldn't give impossible numbers. Therefore, perhaps I made a miscalculation.\n\nWait
Rollout 90, Act: -2.718
16 mod2018⇒ L=2018-1,016=
Rollout 90, Act: -2.840
.\n\n2018*45= 2018*40=80,
Rollout 95, Act: -3.099
for such an answer if it were zero.\n\nSo, clearly, there's an error in this reasoning.\n\n
Rollout 90, Act: -3.212
mod2018. Therefore, L=2018 -1,012=
Rollout 71, Act: -3.281
02=400. then left 4022 -400=36

DOWN_PROJ

Top 16 Positive Activations
Rollout 77, Act: 13.617
*13= 4*91= 364. Yes. So yes, f
Rollout 96, Act: 13.181
�28,561≈3.11\n\nBut precise computation:\n\n28,5
Rollout 38, Act: 13.021
403/217≈1.857x.\n\nComparing with W=(2
Rollout 28, Act: 13.020
��2/3 )2 / (1/18) + Y2/(1/6
Rollout 70, Act: 13.005
0 /7^3=2400/343≈6.997,
Rollout 96, Act: 12.974
= Divide numerator and denominator by 12: 193/169. Similarly,
Rollout 37, Act: 12.623
125 - 27*25 = 1125 - 675
Rollout 62, Act: 12.298
=1100 -10+1=1091. Prime.\n\nHmm. So
Rollout 3, Act: 12.291
0.697 ��28,624; sqrt ��169
Rollout 65, Act: 12.280
/3=3*25*49*361=3*52*7
Rollout 22, Act: 12.238
60539 ~ 0.0255. So total approx 3.3
Rollout 28, Act: 12.235
36k2\n\n-108h -108k\n\n+81=0
Rollout 70, Act: 12.210
(>333), so 7^3=343 (>333, so
Rollout 27, Act: 12.118
94e-26 ��2.228e10 m2/s2.
Rollout 70, Act: 11.988
i=3:1000 -2*343=1000 -6
Rollout 70, Act: 11.986
in base-7: Since 7^3=343, 363-3
Top 16 Negative Activations
Rollout 87, Act: -13.667
\n\nCalculate:\n\n3*log(1 + sqrt(2)) �� 3 * 0.8
Rollout 21, Act: -13.894
=-1, which does not satisfy the equation.\n\nHence such a modified function wouldn’t work as z
Rollout 22, Act: -14.068
1,928 �� 1,777 m/s. Close to 1
Rollout 22, Act: -14.331
* 0.921 = approx 1,770 m/s, which would correspond
Rollout 3, Act: -14.356
016÷0.697≈27,950; sqrt(27
Rollout 81, Act: -14.565
[20/h]/[1 -51/h2] = [20/h]/[(h
Rollout 3, Act: -14.933
/0.697433≈27,950. sqrt(27
Rollout 3, Act: -14.943
/0.697433≈27,950\n\nsqrt(27
Rollout 1, Act: -15.314
balls are not compact hence there is no guarantee.\n\nHences perhaps requires another approach, using the structure of
Rollout 68, Act: -15.631
)=sqrt(254000). Hmmm.\n\nBut regardless, the approximate calculation:\n\nsqrt
Rollout 14, Act: -15.717
right-hand side is 8 times the integral of f squared times integral of x f. Given that integr
Rollout 14, Act: -15.747
some sort of inner product involving x f(x). Hmmm.\n\nWait, what if we make the substitution
Rollout 17, Act: -16.483
by the proton and the probability density |Ψ(r)|2 of the electron in the ground state of a
Rollout 65, Act: -16.562
rest causes a square law.\n\nBut my memory is harkening to a math concept where breaking a rectangle
Rollout 22, Act: -16.671
68.7768 * 157.93. So 18,
Rollout 23, Act: -17.376
1/5.88≈0.1179 s−1\n\nThen lambda_total=

Layer 61

GATE_PROJ

Top 16 Positive Activations
Rollout 66, Act: 10.525
4 and 13: gcd(84,13)=13 factors 13;
Rollout 66, Act: 10.444
2, we get the primitive triplet (84,13,85 ). Thus, check if
Rollout 66, Act: 10.326
be divided by 2 to produce (84,13,85 ). But then how does
Rollout 65, Act: 10.228
, p divides1995*q. Since p and q are coprime, then p divides
Rollout 19, Act: 10.138
17 modk ⇒ 83≡0 modk⇒ thus k divides83.\n\nAs8
Rollout 66, Act: 10.138
down by 2, we have (84,13,85 ). Let's check "
Rollout 0, Act: 10.107
total pairs, so 256/2 =128. Hence 128 such
Rollout 19, Act: 10.107
k. So need x2 ��17 modk. So need congruence to hold for
Rollout 62, Act: 10.106
. But this one is m^2 -m +11. From m=1 to 1
Rollout 62, Act: 10.096
positive integer m such that m^2 -m +11 is a product of at least four not
Rollout 19, Act: 10.080
2.\n\nWe need x2 ��17 mod k.\n\nBut also, note that congruence
Rollout 66, Act: 10.072
the required equation. Hence, perimeter=84 +13 +85=182.
Rollout 65, Act: 10.047
divides1995*q. Since p and q are coprime, then p divides19
Rollout 19, Act: 9.996
4)/2] modk ��10 modk\n\nTherefore, the expression is equivalent to
Rollout 65, Act: 9.994
C2/B <=B,\n\n⇒C2/B <=B,\n\n⇒ C2 <=B2,\n\n⇒
Rollout 19, Act: 9.981
7 modk⇒100≡17 modk⇒ 83≡0modk⇒
Top 16 Negative Activations
Rollout 66, Act: -2.315
[\n \frac{ab}{c - a} = 6p\n \]\n where
Rollout 5, Act: -2.320
+ 25}{b + 75} = \frac{2}{3}\n\]\n
Rollout 27, Act: -2.428
2} m v_{z, \text{rms}}^2\n \]\n - Sol
Rollout 68, Act: -2.431
properties of logarithms:\n\[\n\log(kx) = 2\log(x+2)
Rollout 89, Act: -2.440
\[\n \frac{3}{5}T = 100\n \]\n4
Rollout 59, Act: -2.455
28} + \frac{4}{28} = \frac{10}{28
Rollout 55, Act: -2.477
}{1}.\]Compute $\mathbf{A}^{2018}.$\n<|im_start|>
Rollout 55, Act: -2.491
\(12 \times \frac{\pi}{6} = 2\pi\)), and the
Rollout 72, Act: -2.505
sum is \( 1 + 1 + (-1) = 1 \).\n - This construction
Rollout 53, Act: -2.589
= 2.5\), we find \(AC = 5\).\n\n3. **Setting up
Rollout 5, Act: -2.669
b + 25}{b + 75} = \frac{2}{3}\n\
Rollout 66, Act: -2.680
(84 + 13 + 85) = 6 \times 182
Rollout 35, Act: -2.706
n + 3 \) translates to \( Q(n) = 0 \), meaning \( n_
Rollout 52, Act: -2.733
+ \frac{k(k-1)}{2} = 2014\n\]\nR
Rollout 55, Act: -2.841
12 \times \frac{\pi}{6} = 2\pi\)), and the reflection
Rollout 52, Act: -2.913
n + \frac{k(k-1)}{2} = 2014\n\]\n

UP_PROJ

Top 16 Positive Activations
Rollout 29, Act: 4.803
intuitively, the pressure there might just be rho*g*h. But since the handle has a higher milk
Rollout 29, Act: 4.788
pressure at the bottom of the handle would be rho*g*H, but if the main jug is connected
Rollout 29, Act: 4.682
is rho*g*h and the handle's is rho*g*H. But the problem is asking about the
Rollout 29, Act: 4.649
's bottom (if considered separately) would have rho*g*H. But the question is about the gauge
Rollout 29, Act: 4.590
of the main jug would be atmospheric pressure + rho*g*h. Since it's open, gauge pressure is
Rollout 29, Act: 4.543
pressure at the bottom of the main jug is rho*g*h. However, maybe because the handle's milk
Rollout 29, Act: 4.514
main jug's bottom has a gauge pressure of rho*g*h, while the handle's bottom (if considered
Rollout 29, Act: 4.506
pressure of the sealed air in the handle plus rho*g*H. Since the pressure at the interface between
Rollout 29, Act: 4.503
*H. Then the gauge pressure would be rho*g*H - atmospheric pressure. But since the main
Rollout 25, Act: 4.475
. Similarly, acceleration on Earth is F/M = G*m / d^2, which would cause Earth
Rollout 29, Act: 4.464
equal:\n\nrho*g*h + atmospheric pressure = rho*g*H + P_handle\n\nTherefore, P_handle
Rollout 29, Act: 4.440
pressure at the bottom of the handle would be rho*g*H. Then the gauge pressure would be rho
Rollout 29, Act: 4.426
atmospheric, would be (atmospheric + rho*g*h) - atmospheric = rho*g*h in the
Rollout 29, Act: 4.384
jug, which is h, so P = rho*g*h. However, if the handle is somehow still
Rollout 48, Act: 4.380
valid, indicating no error in the header.\n\n\boxed{No}<|im_end|>
Rollout 29, Act: 4.343
For the main jug: P_gauge = rho*g*h\n\nFor the handle: P_gauge =
Top 16 Negative Activations
Rollout 17, Act: -7.581
/s2, but here in any case, the units need to resolve, but since the previous calculation showed
Rollout 30, Act: -7.597
? Wait, maybe not. It depends if there is a miscalculation here.\n\nLet me check
Rollout 73, Act: -7.612
quickly, but modulo 1000 could help.\n\nWait, maybe verifying N=163
Rollout 33, Act: -7.621
-48 (m=3). Let me check:\n\n720 - 372 =
Rollout 73, Act: -7.630
16398. Alternatively, perhaps I made a mistake in calculation.\n\nEarlier steps:\n\nWe had
Rollout 16, Act: -7.641
4818...\n\nWait, but perhaps that's getting too approximate. Let's compute each fraction exactly
Rollout 30, Act: -7.677
Wait, maybe not. It depends if there is a miscalculation here.\n\nLet me check through
Rollout 93, Act: -7.716
/N)]^{2N}. Wait, is there a known formula for this kind of sum?\n\nAlternatively,
Rollout 2, Act: -7.723
izing each term, might use modulus1000. However, since 1000=
Rollout 48, Act: -7.755
. So F946).\n\nBut wait, maybe handling it in hex:\n\n0x450
Rollout 76, Act: -7.771
But that might be complicated. Alternatively, perhaps there's a mistake in inclusion-exclusion.\n\nAlternatively, think
Rollout 73, Act: -7.806
0=398.\n\nBut might we have made a miscalculation when computing sum_{i
Rollout 48, Act: -7.819
x10000. But I can do the sum:\n\n4500\n\n+B6
Rollout 89, Act: -7.874
must be a whole number, so seems like there is inconsistency here.\n\nWait, but maybe the problem is
Rollout 72, Act: -7.881
. Going through primes, but maybe we can find a pattern.\n\nSo so far, n=2,
Rollout 33, Act: -7.887
) -48 (m=3). Let me check:\n\n720 - 372

DOWN_PROJ

Top 16 Positive Activations
Rollout 65, Act: 13.694
1995, k*c].\n\nPossibly re-ordered:\n\nSuppose original ordered [a,
Rollout 48, Act: 13.090
=3,609 rem 14 → E. 3,609 /1
Rollout 91, Act: 13.026
the value of \(|\cot B - \cot C|\) is \(\boxed{2}\).
Rollout 25, Act: 12.721
1.008e13\n\nNumerator: 1.008e1
Rollout 6, Act: 12.645
log n + n^k �� m? Unlikely. For example, m is dominated by n
Rollout 81, Act: 12.460
1/h2 )\n\n Withh=11:\n\nNume:20/11.\n\n Denom
Rollout 81, Act: 12.315
, no.\n\nWait, sinA=2Δ/(bc).\n\ncosA=(b2 +c2
Rollout 81, Act: 12.304
sinA.\n\n thus sinA = 2Δ/(bc).\n\nLaw of Cosines: a2=
Rollout 23, Act: 12.262
But ln(1.9e31)=nearly 72. Therefore, exp(7
Rollout 37, Act: 12.227
! as a factor. But I cannot explain it concretely.\n\nAlternatively, consider that the determinant can
Rollout 84, Act: 12.116
Multiply through by 18: 9 - 18x = 4*(1 - x
Rollout 22, Act: 11.880
.002016 kg/mol\n\nNumerator: 8 * R * T =8
Rollout 16, Act: 11.876
which might be complicated. However, maybe we can reindex it. Let m = k -1.
Rollout 11, Act: 11.871
\overline{ab} + \overline{bc} + \overline{ca} =
Rollout 81, Act: 11.745
121/11≈11。\n\n Confusion again!\n\n Thus Two different equations arise:\n\nFrom
Rollout 25, Act: 11.369
.49e6 m2/s2\n\nNumerator: 10.08e1
Top 16 Negative Activations
Rollout 33, Act: -17.215
km - km-1)! * (6 - km)!.\n\nHence, for inclusion-exclusion,
Rollout 36, Act: -17.253
2 = 18000 -- (2')\n\nEquation (3): S - 2
Rollout 36, Act: -17.315
us subtract equations (1'), (2'), (3') to find relationships.\n\nFrom (1') -
Rollout 0, Act: -17.404
20! or a < sqrt(20!). Therefore, the number of such a is half
Rollout 36, Act: -17.659
4')\n\nLet us subtract equations (1'), (2'), (3') to find relationships.\n\nFrom (
Rollout 33, Act: -18.348
k4 -k3)!*(6 -k4)!.\n\nIn other words, each consecutive difference.
Rollout 36, Act: -18.703
as before)\n\nSimilarly, (1') - (3'): -2s x + 2s z
Rollout 36, Act: -18.811
= -864 => substitute into Equation (1'):\n\nS + (-864) =
Rollout 24, Act: -19.049
, a(r, c) ≤ a(r, c').\n\nSimilarly, From the saddle pair (R
Rollout 36, Act: -19.204
- s2\n\nSo substituting into equation (4'):\n\n36000 + 2
Rollout 36, Act: -19.694
, if we subtract equations (2') - (3'): -2s y + 2s z
Rollout 36, Act: -20.356
from an equation (1').\n\nFrom equation (1'):\n\nS = 36000
Rollout 36, Act: -21.008
'):\n\nEquation (d) + equation (1'):\n\n(S - 2s x + s
Rollout 36, Act: -21.417
= -864, add to equation (1'):\n\nEquation (d) + equation (
Rollout 24, Act: -21.794
, a(r, c) ≤ a(r, c').\n\nThen, for (R, C)
Rollout 36, Act: -25.594
to find relationships.\n\nFrom (1') - (2'): -2s x + 2s y

Layer 62

GATE_PROJ

Top 16 Positive Activations
Rollout 35, Act: 0.802
) are roots of \( Q(n) \).\n\nGiven \( P(17) = 10
Rollout 93, Act: 0.760
+ e^{-i\theta}}{2}\right)^{2014} = \frac
Rollout 35, Act: 0.709
24) = 17 \), we find:\n\[\nQ(17) =
Rollout 90, Act: 0.563
and} \quad B \equiv 0 \pmod{2018}\n \]\n
Rollout 17, Act: 0.519
probability density at the origin is \(|\Psi(0)|^2 = \frac{1}{\
Rollout 20, Act: 0.488
\( p^{kn} \equiv 1 \pmod{47} \). This requires \(
Rollout 19, Act: 0.426
frac{n(n+1)}{2} \right)^2 \equiv 17 \pmod
Rollout 36, Act: 0.371
8064}{2s} \right)^2\n\]\n\nAfter simplification and solving
Rollout 59, Act: 0.369
= \left(\frac{1}{2}\right)^2 \cdot \frac{3}{7
Rollout 8, Act: 0.331
$5$ of the cards in a row in a random order. She will be happy if and only
Rollout 59, Act: 0.322
\left(\frac{4}{28}\right)^2 \\\n&= \frac{9 +
Rollout 4, Act: 0.312
^2}{(C_1 + C_2)^2} \quad \text{and}
Rollout 4, Act: 0.288
C_1 + C_2}\n\]\n\nGiven \( C_1 = 2000
Rollout 60, Act: 0.280
Hz}\n\]\n\nThe beat frequency is the difference between the reflected frequency \( f'' \) and the
Rollout 19, Act: 0.240
)(k - 4)}{2} \right)^2 \equiv 17 \pmod
Rollout 5, Act: 0.213
b = 175 \). Substituting these values:\n\[\nx = \sqrt{\
Top 16 Negative Activations
Rollout 70, Act: -7.922
2d1 >=157 =>d1 >=78.5 But d1<=6,
Rollout 90, Act: -7.925
1,928=2018×5000 -8,072
Rollout 75, Act: -7.939
a2 +b2 +c2 -3)/4. But a,b,c are odd, squares
Rollout 75, Act: -7.953
2k +1)2 =4k2 +4k +1=4k(k +1
Rollout 86, Act: -7.962
/9=7/9≈0.777.\n\nTherefore, integral case7≈0.
Rollout 45, Act: -7.976
is f(x)=x3 +4x2 -4. But check if with this, do the
Rollout 82, Act: -7.986
8=80/38≈2.105. So in this case, the binary
Rollout 51, Act: -7.999
, maybe it's positive 4/63 sin8θ? Let me check... Because sometimes during
Rollout 65, Act: -8.016
the number of divisors is (1+1)^4=16.\n\nBut But. However,
Rollout 70, Act: -8.027
2k ��3 mod5, so k≡4 mod5 (since 2*4=
Rollout 20, Act: -8.031
kn})=kn +1≡1 when k≡0 mod43. But the problem requires for
Rollout 36, Act: -8.039
764; =>x≈2.0105...\n\nBut likely, sqrt(36
Rollout 65, Act: -8.045
The number of divisors is (2+1)^4=8l. All the divisors a
Rollout 75, Act: -8.065
the resulting 2k= (S -3)/4=2*(a2 +b2 +
Rollout 68, Act: -8.137
(-4))x+ 4= x2 +8x +4=0. Then:\n\nDisc
Rollout 45, Act: -8.154
m)x2 - (5m +1)x -4 -4m ≥0\n\nOr,\n\nx

UP_PROJ

Top 16 Positive Activations
Rollout 94, Act: 1.312
c), and (d), and the answer choices are combinations of these. First, let me recall what
Rollout 42, Act: 1.296
cyclopropane ring. That might lead to a conjugated diene or a larger carbocycle
Rollout 49, Act: 1.276
, the perfect subset property holds? In Polish spaces, if it's an uncountable Polish space,
Rollout 1, Act: 1.273
:\n\n-2〈x. This action would be the same as above for linear equations, but we already
Rollout 42, Act: 1.272
bond, leading to a more conjugated structure. The main product would likely have the most stable alkene
Rollout 42, Act: 1.251
carbocation would create a larger ring.\n\nAlternatively, a hydride shift could occur from the cyclo
Rollout 77, Act: 1.206
9). Since 266=14*19. Therefore 14*19
Rollout 42, Act: 1.206
spiro structure.\n\nAlternatively, let's think of the result. If (D) spiro[3
Rollout 54, Act: 1.200
base.\n\nAlso, propionate (3) is a very weak base (since propionic acid is a
Rollout 26, Act: 1.199
is two. The result would be 303.4, which has three sig figs.
Rollout 25, Act: 1.180
67e-11\n\n= 6.13e14 /6.67
Rollout 24, Act: 1.169
leqslant a\left(r, c^{\prime}\right)$ for all $r \
Rollout 97, Act: 1.162
^{-1.\n\nPer the IUPAC definition, reaction rate (r) is defined as rate= rate
Rollout 86, Act: 1.137
pi/2)=pi/16≈0.19635 ��0.1
Rollout 54, Act: 1.133
3):\n\nPropionate (RCOO−) is the weakest base, so in polar protic solution,
Rollout 42, Act: 1.102
ed and one four-membered ring.\n\nAlternatively, the formation of a spiro compound might not even require
Top 16 Negative Activations
Rollout 41, Act: -6.442
27) seems correct, unless there is a miscalculation. Then I can search the current
Rollout 79, Act: -6.474
anagram. So "Minor changes" can be rearranged. Let's try to split into two parts
Rollout 14, Act: -6.488
\frac{f(x)}{\sqrt{x f(x)}}}$, but that feels like a stretch and
Rollout 79, Act: -6.488
fits 8,4 is "CHANGING HANS". However, after reconsidering and in fact
Rollout 78, Act: -6.518
22, n = 27, so m + n = 49. The answer is
Rollout 78, Act: -6.541
22 and n is 27, so m + n = 22 + 27
Rollout 89, Act: -6.554
.\n\nTherefore, unless problem is altered, thus:\n\nSuppose there's misprint in the problem. Instead
Rollout 30, Act: -6.581
, maybe not. It depends if there is a miscalculation here.\n\nLet me check through each
Rollout 34, Act: -6.603
+ C.\n\nThus:\n\n1 = 2 + C => C = -1\n\nTherefore, the solution
Rollout 71, Act: -6.631
Then |a| + |b| + |c|. \n\nGiven that, maybe another approach: with
Rollout 36, Act: -6.659
perfect square, but our calculation might have been error-prone.\n\nAlternatively, maybe there is an arithmetic mistake
Rollout 21, Act: -6.677
any real number x, there's an integer k s.t. x ∈ [k, k+1
Rollout 67, Act: -6.718
3721d2.\n\nSo sqrt(D) = sqrt(3721d2
Rollout 79, Act: -6.759
(8?) Let me check:\n\n INCIDENT → I-N-C-I-D-E-N-T – 8 letters
Rollout 79, Act: -6.937
4, but if "Homing" and "Cranes"? No. Hmm.\n\nAnother approach:
Rollout 79, Act: -7.040
word.\n\nAlternatively, "Machines" and "Rong"? Not likely.\n\nAnother angle: "Mech

DOWN_PROJ

Top 16 Positive Activations
Rollout 30, Act: 28.054
1\right)$ and $b_{n+1}=a_{n} b_{n-1
Rollout 16, Act: 24.727
,2)=15, (-1)^{3}=-1, denominator 1 -1/
Rollout 5, Act: 24.372
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 47, Act: 24.372
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 60, Act: 24.372
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 94, Act: 24.372
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 34, Act: 24.372
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 55, Act: 24.372
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 43, Act: 24.372
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 78, Act: 24.372
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 63, Act: 24.372
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 84, Act: 24.372
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 88, Act: 24.370
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 14, Act: 24.357
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 15, Act: 24.357
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Rollout 16, Act: 24.357
<|im_start|>system\nYou are a helpful mathematics assistant.\n<|im_start|>user
Top 16 Negative Activations
Rollout 52, Act: -14.430
2013 with 4028 terms sum to result.\n\nBut -2013
Rollout 12, Act: -14.469
- t)(t + a)(t + b)} } \n\nBut maybe that's diverging.\n\nAlternatively
Rollout 30, Act: -14.490
} -1/a_{n+3} )\n\nWe tried expressing each term, and ended up finding that
Rollout 79, Act: -14.627
possible letter combinations.\n\nH, O, M, I, N, G (homing). Then remaining
Rollout 75, Act: -14.632
into two squares each, but for 32 it can't be, but in that case, is
Rollout 75, Act: -14.671
squares. Thus, for S=35, it works. Similarly, for other cases?\n\nFor example
Rollout 71, Act: -14.862
3d2=2011, given it's prime, the only solutions are (±4
Rollout 75, Act: -15.164
each 9 is a single square. But four squares:\n\n16 +9 +9 +1 =
Rollout 39, Act: -15.454
9...9997101, it can be split as 999...9
Rollout 11, Act: -15.458
work. Trying a specific example seems not possible as it resulted in prohibited things. Hmmm.\n\nWait,
Rollout 65, Act: -15.634
1975, c.\n\nWhich has ratiostroubled as follows:\n\n original prism 35
Rollout 95, Act: -15.652
equal areas, thus the problem's conditions are incorrect(but The problem says that it can be split in
Rollout 66, Act: -17.928
we end up with both legs even, which confilcts with the primitive triple. So this suggests that
Rollout 93, Act: -17.948
)^k is zero at x=x_k. Hmmm.\n\nBut x_k corresponds to k from1
Rollout 49, Act: -23.696
is at most 2^{2^{d(X)}}, where d(X) is the density. But
Rollout 12, Act: -26.223
\nu^2)(1 - t^2)}}\n\]\n\nNext, we recognize this integral as

Layer 63

GATE_PROJ

Top 16 Positive Activations
Rollout 89, Act: 7.290
result in fractions of people would indicate a miscalculation. While this problem produces 66.(
Rollout 45, Act: 7.133
+ b + 2c + 4\n\nSimplify: 20 = 6 +
Rollout 25, Act: 7.109
). So in real case, the trajectory is hyperbola with perigee r_p and the ap
Rollout 52, Act: 7.095
= -33.5. Not integer. Disregard.\n\n9. k=212
Rollout 68, Act: 7.062
2 + 4x + 4. Rearranging terms, we get:\n\nx^2 +
Rollout 14, Act: 7.052
bit tricky. Maybe express this as:\n\nRearranged,\n\n$$\nf(x) = a -
Rollout 19, Act: 7.000
-5= -4 which isn’t positive. Disregard. k=83, which gives
Rollout 88, Act: 6.997
licative parameters, biases are additive. So maybe the question is only asking about the weights. Then it would
Rollout 8, Act: 6.995
be adjacent). But the way the problem is phrased: "all the red cards laid out are
Rollout 89, Act: 6.946
- (2/5)*T, right?\n\nSimplifying that, boys = (1 -
Rollout 89, Act: 6.909
that others may got.\n\nAlternatively, perhaps my overcomplicating.\n\nWait, is it possible in non
Rollout 25, Act: 6.872
2/2, but since it's a hyperbolic trajectory, ε is positive.\n\nThen, we
Rollout 42, Act: 6.844
very unstable. So perhaps it undergoes a hydride shift? Let's see.\n\nWait, when
Rollout 70, Act: 6.821
1*7 + d_0).\n\nRearranging,\n\nd_k*(10^k -
Rollout 44, Act: 6.805
the irreversible process. Wait, but that seems counterintuitive, because in irreversible processes there are losses.
Rollout 81, Act: 6.789
BAD and CAD are acute, so sum can be acute or possibly greater than 90, but given
Top 16 Negative Activations
Rollout 46, Act: -0.471
gets complicated.\n\nAlternatively, the question is asking for the Laplacian of U, but Laplace's
Rollout 59, Act: -0.486
Phil have two fair coins and a third coin that comes up heads with probability $\frac47$ .
Rollout 87, Act: -0.518
After first integrating with respect to x:\n\nThe integral ��0^a dx / sqrt(x2 +
Rollout 46, Act: -0.518
L/∂x = �� v_y + ��U/∂x\n\nTherefore, equation:\n\n
Rollout 49, Act: -0.526
X. A Hausdorff space with Gδ diagonal is first-countable. Ah,
Rollout 57, Act: -0.548
of its points.\n\nHowever,the Cantor set C has measure zero. So,every neighbourhood of a point
Rollout 46, Act: -0.565
/∂x = �� dy/dt - ��U/∂x\n\nTherefore, putting into
Rollout 3, Act: -0.566
\) is the molar mass of radon in kg/mol.\n\n**Steps:**\n\n1. **
Rollout 60, Act: -0.568
\mathrm{~ms}^{-1}$. The reflected signal is received by the source and superposed
Rollout 49, Act: -0.608
σ-compact.\n\nHence, X is ���-dimensional? Maybe not necessarily.\n\nAlternatively, embed
Rollout 80, Act: -0.628
surface.\n\nIn the movie 2012, neutrinos have mutated and now are completely absorbed by
Rollout 47, Act: -0.646
sides and the walls of the tunnel. Also, the air flow with respect to the train is steady and
Rollout 97, Act: -0.661
atm.\n\nUnder stoiechmentric decomposition, y is the extent of reaction in terms of how many
Rollout 97, Act: -0.702
if.\n\nAssumption: if N2O5 follows P(t}= exp(-kt}., then
Rollout 44, Act: -0.900
1}$ to $\mathrm{V}_{2}$ under adiabatic conditions as compared to that when
Rollout 13, Act: -0.987
is $12\%$. She makes no payments for 4 years, but has to pay

UP_PROJ

Top 16 Positive Activations
Rollout 40, Act: 4.885
, the term was 10/(1 - ��)*(1 + i). Wait, earlier:\n\n
Rollout 34, Act: 4.798
�� e^{2s} ds from 0 to t is (1/2)(e^{
Rollout 40, Act: 4.783
1 )\n\nHence, 1/(1 - ��) = [1 + i(√2
Rollout 40, Act: 4.773
^n z0 + b (sum_{k=0}^{n-1} a^k ).
Rollout 8, Act: 4.716
k) = 5! / (5 - k)!.\n\nSimilarly, arrange (5 -k)
Rollout 40, Act: 4.595
�2\n\nTherefore, 1/(1 - ��) = [ (1 - ��2
Rollout 23, Act: 4.524
147 / 0.098 �� 7.073... min−
Rollout 69, Act: 4.514
38 / 0.9848 �� 0.00343.\n\n
Rollout 8, Act: 4.482
(n,k) = n! / (n - k)!.\n\nSimilarly, for a different color sequence with
Rollout 92, Act: 4.465
i = 200/-123 �� -1.6260...\n\nThus
Rollout 89, Act: 4.453
2/5 *500/3 ��200/3≈66.
Rollout 40, Act: 4.397
Then plugged into term [10/(1 - �� )]*(1 +i). Then multiplying out
Rollout 58, Act: 4.390
total number of questions to be the sum from 1 to 2018. But just to
Rollout 13, Act: 4.385
0*(1 + 0.12/4)^(4*4)\n\nFirst, calculate
Rollout 44, Act: 4.380
γ-1)= T2*(20/1)^(2/3). But T2=
Rollout 92, Act: 4.371
1/(-0.615) ��5.405 -1.6
Top 16 Negative Activations
Rollout 10, Act: -2.884
bbles. The key is to identify losing positions (P-positions) where every move leads to a
Rollout 87, Act: -2.887
constant involving logarithmic terms. After evaluating this integral (a known result in electromagnetism), the potential
Rollout 98, Act: -2.888
** \n - A white dwarf progenitor with **0.85Msun** is inconsistent:
Rollout 42, Act: -2.894
ocation rearrangement** occurs. This rearrangement involves **ring-opening of the cyclopropane** to
Rollout 82, Act: -2.895
- Each group is encoded with a binary prefix code (0 or 1), resulting in the following cod
Rollout 3, Act: -2.901
ounding to Significant Figures:**\n Given the inputs (molar mass \(222 \, \
Rollout 22, Act: -2.966
gas (H2) at standard temperature and pressure (STP), we use the formula for the mean
Rollout 48, Act: -2.987
**Total Length:** `0034` (52 bytes total packet length).\n - **
Rollout 6, Act: -2.998
.\n\nAnother angle: Let S_L contain the tuple <n, α_n >. But encode this in
Rollout 10, Act: -3.014
positions) where every move leads to a winning position (N-position) for the opponent.\n\n1. **
Rollout 98, Act: -3.033
un) with long lifetimes. Their lifetimes (10 Gyr, ~24 Gyr
Rollout 22, Act: -3.080
- \( T \) is the temperature in Kelvin (assumed to be 273.1
Rollout 10, Act: -3.151
**:\n - Players take turns removing square numbers (1, 4, 9, 1
Rollout 65, Act: -3.156
995 itself. Half of these divisors (excluding the midpoint) are less than 19
Rollout 6, Act: -3.485
oracle can contain those positions. So, for each (n, i), where i is a bit position
Rollout 48, Act: -3.543
4 (IPv4), IHL is 5 (20 bytes).\n - **Type of Service

DOWN_PROJ

Top 16 Positive Activations
Rollout 79, Act: 33.036
thinking, the intended answer is **MONITOR'S GAZE**, but the definitive answer, following standard crossword
Rollout 79, Act: 30.549
.\n\nAnother idea: The solution is "MACHINE GUNS". "Machine" (7 letters) and
Rollout 95, Act: 25.314
0为1,每个 trapezoidal region ���要s1, s2等。\n\n
Rollout 14, Act: 23.731
g(y). Then J scales with d^2 ��g^2 dx= d^2/c
Rollout 79, Act: 23.450
S. Then 4-letter word could be "gros" — no. Not plausible.\n\nAlternatively,
Rollout 9, Act: 23.330
$ units. Isosceles triangle $GEM$ has base $EM$ , and the
Rollout 81, Act: 22.441
。 But as angle BAC is required to be ��≈72 degrees.\n\n which contradicts because
Rollout 49, Act: 22.364
} at x. For each n, U_n �� A is uncountable. Let’s choose
Rollout 44, Act: 21.677
adiabatic conditions with $\mathrm{T}_{1} \neq \mathrm{T}_{2}
Rollout 79, Act: 21.429
don't have. \n\nScraps: "Morning GCA" — nothing. \n\nOR some terms like
Rollout 82, Act: 21.426
但霍夫曼算法通常会倾向让较��概率符号用较短的代码。但是在
Rollout 57, Act: 20.934
x处为1,并且A⊆P?\n\n ���而,因为无理数的集合
Rollout 56, Act: 20.563
a general formula.\n\nFirst triangle: S0 = ��3 /2 ~0.866
Rollout 14, Act: 20.207
(�� f(x) dx )^3 = �� f(x) dx * �� f(y
Rollout 14, Act: 20.193
��0^∞ f(x) dx = ��0^∞ f(x) \cdot1
Rollout 33, Act: 20.058
permutation. Which matches f(2)=1.\n\nForn=3 as earlier,it's3.\n\nLet
Top 16 Negative Activations
Rollout 60, Act: -20.011
the Doppler effect formula for a moving source:\n\[\nf'' = f' \frac{v
Rollout 48, Act: -20.267
4\n3. B612 = 0xB612\n4. 40
Rollout 55, Act: -20.353
}\), resulting in:\n \[\n \begin{pmatrix} \frac{1}{2
Rollout 79, Act: -21.079
-letter words from these letters include:\n\n- "Hernias" plus some others. No. Maybe "
Rollout 26, Act: -21.338
py change when going from Hg(l) to Hg(g), which is the enthalpy of
Rollout 48, Act: -21.642
63,814 in hex is 0xF946. Wait, this is a
Rollout 48, Act: -21.699
0x4000. So 0xFB46 +0x4000
Rollout 32, Act: -21.747
ω.\n\nSo M2 is:\n\n\[\n\begin{pmatrix} 0 & 1 \\
Rollout 48, Act: -21.918
0 to 0xFFFF?), wait no, 0xFB46 is a 16-bit value
Rollout 55, Act: -22.103
\frac{1}{2} \end{pmatrix}\n\]\n\nYep, so that looks about
Rollout 49, Act: -22.666
endixson derivative.\n\nThe Cantor-Bendixson theorem states that in a separable metr
Rollout 11, Act: -24.030
} = 1/(\overline{ab})). Therefore, maybe expressions can be written in terms
Rollout 32, Act: -24.282
omega \\ 1 & 0 \end{pmatrix}.\]Find the sum of the entries
Rollout 48, Act: -24.371
0x4000: sum becomes 0xFB46 + 0x400
Rollout 17, Act: -26.246
magnitude of the integral \(|\mathbf{I}|\) is \(\boxed{2.5
Rollout 17, Act: -26.879
magnitude of the integral \(|\mathbf{I}|\) for the given problem, we need to